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\begin{center}
\vskip 1cm{\LARGE\bf New Estimates for the $n$th Prime Number}
\vskip 1cm
\large
Christian Axler\\
Department of Mathematics\\
Heinrich-Heine-University\\
40225 D\"usseldorf\\
Germany\\
\href{mailto:Christian.Axler@hhu.de}{\tt Christian.Axler@hhu.de}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper we establish new upper and lower bounds for the $n$th
prime number $p_n$, which improve several existing bounds of similar shape. As
the main tool, we use some explicit estimates recently obtained for the
prime counting function. A further main tool is the use of estimates
concerning the reciprocal of $\log p_n$. As an application, we derive
new estimates for $\vartheta(p_n)$, where $\vartheta(x)$ is Chebyshev's
$\vartheta$-function.
\end{abstract}

\section{Introduction}

Let $p_n$ denote the $n$th prime number and let $\pi(x)$ be the number of primes not exceeding $x$. In 1896, Hadamard \cite{hadamard1896} and de la 
Vall\'{e}e-Poussin \cite{vallee1896} independently proved the asymptotic formula $\pi(x) \sim x/\log x$ as $x \to \infty$, which is known as the \textit{prime 
number theorem}. (Here $\log x$ is the natural logarithm of $x$.) As a consequence of the prime number theorem, one gets the asymptotic expression 
\begin{equation}
p_n \sim n \log n \q\q (n \to \infty). \tag{1.1} \label{1.1}
\end{equation}
Here $p_n$ is the $n$th prime. Cipolla \cite{cp} found a more precise result. He showed that for every positive integer $m$ there exist unique monic 
polynomials $T_1, \ldots, T_m$ with rational coefficients and $\deg(T_k)= k$ \nolinebreak with
\begin{equation}
p_n = n \left( \log n + \log \log n - 1 + \sum_{k=1}^m \frac{(-1)^{k+1}T_k(\log \log n)}{k\log^kn} \right) 
+ O\left( \frac{n (\log \log n)^{m+1}}{\log^{m+1}n} \right). \tag{1.2} \label{1.2}
\end{equation}
The polynomials $T_k$ can be computed explicitly. In particular, $T_1(x) = x - 2$ and $T_2(x) = x^2 - 6x + 11$ (see Cipolla \cite{cp} or Salvy \cite{salvy} for 
further details). Since the computation of the $n$th prime number is difficult for large $n$, we are interested in explicit estimates for $p_n$. The 
asymptotic formula \eqref{1.2} yields
\begin{align}
&amp; p_n &gt; n \log n, \tag{1.3} \label{1.3} \\
&amp; p_n &lt; n ( \log n + \log \log n), \tag{1.4} \label{1.4} \\
&amp; p_n &gt; n ( \log n + \log \log n - 1) \tag{1.5} \label{1.5}
\end{align}
for all sufficiently large values of $n$. The first result concerning a lower bound for the $n$th prime number is due to Rosser \cite[Theorem 1]{jr}. He 
showed that the inequality \eqref{1.3} holds for every positive integer $n$. In the literature, this result is often called \textit{Rosser's theorem}. 
Moreover, he proved \cite[Theorem 2]{jr} that
\begin{equation}
p_n &lt; n(\log n + 2 \log \log n) \tag{1.6} \label{1.6}
\end{equation}
for every $n \geq 4$. The next results concerning the upper and lower bounds that correspond to the first three terms of \eqref{1.2} are due to Rosser and 
Schoenfeld \cite[Theorem 3]{rosserschoenfeld1962}. They refined Rosser's theorem and the inequality \eqref{1.6} by showing that
\begin{displaymath}
p_{n} &gt; n(\log n + \log \log n - 1.5) 
\end{displaymath}
for every $n \geq 2$ and that the inequality
\begin{equation}
p_{n} &lt; n(\log n + \log \log n - 0.5) \tag{1.7} \label{1.7}
\end{equation}
holds for every $n \geq 20$. The inequality \eqref{1.7} implies that \eqref{1.4} is fulfilled for every $n \geq 6$. Based on their estimates for the Chebyshev 
functions $\psi(x)$ and $\vartheta(x)$, Rosser and Schoenfeld \cite{rosserschoenfeld1975} announced to have new estimates for the $n$th prime number $p_n$ but 
they have never published the details. In the direction of \eqref{1.5}, Robin \cite[Lemme 3, Th\'{e}or\`{e}me 8]{robin1983} showed that
\begin{equation}
p_{n} \geq n(\log n + \log \log n - 1.0072629) \tag{1.8} \label{1.8}
\end{equation}
for every $n \geq 2$, and that the inequality \eqref{1.5} holds for every integer $n$ such that $2 \leq n \leq \pi(10^{11})$. Massias and Robin 
\cite[Th\'{e}or\`{e}me A]{mr} gave a series of improvements of \eqref{1.7} and \eqref{1.8}. For instance, they have found that $p_{n} \geq n(\log n + \log \log 
n - 1.002872)$ for every $n \geq 2$. Dusart \cite[p.\:54]{pd} showed that the inequality 
\begin{equation}
p_n \leq n \left( \log n + \log \log n - 1 + \frac{\log \log n - 1.8}{\log n} \right) \tag{1.9} \label{1.9}
\end{equation}
holds for every $n \geq 27\,076$. Further, he \cite[Theorem 3]{dusart1999} made a breakthrough concerning the inequality \eqref{1.5} by showing that 
this inequality holds for every $n \geq 2$. The current best estimates for the $n$th prime, which correspond to the first terms in \eqref{1.2}, are also given 
by Dusart \cite[Propositions 5.15 and 5.16]{dusart2017}. He used explicit estimates for Chebyshev's $\vartheta$-function to show that the inequality
\begin{equation}
p_n \leq n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} \right), \tag{1.10} \label{1.10}
\end{equation}
which corresponds to the first four terms of \eqref{1.2}, holds for every $n \geq 688\,383$ and that
\begin{equation}
p_n \geq n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2.1}{\log n} \right) \tag{1.11} \label{1.11}
\end{equation}
for every $n \geq 3$. The goal of this paper is to improve the inequalities \eqref{1.10} and \eqref{1.11} with regard to Cipolla's asymptotic expansion 
\eqref{1.2}. For this purpose, we use estimates for the quantity $1/\log p_n$ and some estimates \cite{ca2017} for the prime counting function $\pi(x)$ to 
obtain the following refinement of \eqref{1.10}.

\begin{theorem} \label{thm101}
For every integer $n \geq 46\,254\,381$, we have
\begin{equation}
p_n &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n + 10.667}{2 \log^2 n} \right). \tag{1.12} 
\label{1.12}
\end{equation}
\end{theorem}

Under the assumption that the Riemann hypothesis is true, Dusart \cite[Theorem 3.4]{dusart2018} found that
\begin{equation}
p_n &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n}{2 \log^2 n} \right). \tag{1.13} 
\label{1.13}
\end{equation}
for every integer $n \geq 3468$. Using Theorem \ref{thm101} and a computer for smaller values of $n$, we get

\begin{corollary} \label{kor102}
The inequality \eqref{1.13} holds unconditionally for every $n \geq 3468$.
\end{corollary}

In the other direction, we find the following result which yields a lower bound for the $n$th prime number in a bounded range.

\begin{theorem} \label{thm103}
For every integer $n$ satisfying $2 \leq n \leq \pi(10^{19}) = 234\,057\,667\,276\,344\,607$, we have
\begin{displaymath}
p_n &gt;  n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2 - 6\log \log n + 11.25}{2 \log^2 n} \right).
\end{displaymath}
\end{theorem}

Finally, we use Theorem \ref{thm103} to give the following improvement of \eqref{1.11}.

\begin{theorem} \label{thm104}
For every integer $n \geq 2$, we have
\begin{equation}
p_n &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log\log n)^2-6\log \log n + 11.321}{2\log^2 n} \right). \tag{1.14} 
\label{1.14}
\end{equation}
\end{theorem}

We get the following corollary which was already known under the assumption that the Riemann hypothesis is true (see Dusart \cite[Theorem 3.4]{dusart2018}).

\begin{corollary} \label{kor105}
For every $n \geq 2$, we have
\begin{displaymath}
p_n &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log\log n)^2}{2\log^2 n} \right).
\end{displaymath}
\end{corollary}

In Section 6 we apply the Theorems \ref{thm101} and \ref{thm104} to find some refined estimates for $\vartheta(p_n)$, where $\vartheta(x) = \sum_{p 
\leq x} \log p$ is Chebyshev's $\vartheta$-function.

\begin{notation}
Throughout this paper, let $n$ denote a positive integer. For better readability, in the majority of the proofs we use the notation
\begin{displaymath}
w = \log \log n, \q y = \log n, \q z = \log p_n.
\end{displaymath}
\end{notation}

\section{Effective estimates for the reciprocal of $\log p_n$}

Let $m$ be a positive integer. 
Using Panaitopol's asymptotic formula for the 
prime counting function $\pi(x)$ --- see \cite{pana} --- we see that
\begin{equation}
p_n = n \left( \log p_n - 1 - \frac{1}{\log p_n} - \frac{3}{\log^2 p_n}  - \cdots - \frac{k_m}{\log^m p_n} \right) + O \left( \frac{n}{\log^{m+1}n} \right), 
\tag{2.1} \label{2.1}
\end{equation}
where the positive integers $k_1, \ldots, k_m$ are given by the recurrence formula
\begin{displaymath}
k_m + 1!k_{m-1} + 2!k_{m-2} + \cdots + (m-1)!k_1 = m \cdot m!.
\end{displaymath}
So, in order to prove Theorems \ref{thm101} and \ref{thm104}, we first use some results of \cite{ca2017} concerning effective estimates for $\pi(x)$ which 
imply estimates for the $n$th prime number $p_n$ in the direction of \eqref{2.1}. Then we apply the estimates for the quantity $1/\log p_n$ obtained in this 
section. Cipolla \cite[p.\:139]{cp} showed that
\begin{displaymath}
\frac{1}{\log p_n} = \frac{1}{\log n} - \frac{\log \log n}{\log^2n} + o \left( \frac{1}{\log^2 n} \right).
\end{displaymath}
Concerning this asymptotic formula, we give the following inequality involving $1/\log p_n$, where the polynomials $P_1, \ldots, P_4 \in \Z[x]$ are given by
\begin{align*}
P_1(x) &amp; = 3x^2 - 6x + 5, \\
P_2(x) &amp; = 5x^3 - 24x^2 + 39x - 14, \\
P_3(x) &amp; = 7x^4 - 48x^3 + 120x^2 - 124x + 51, \\
P_4(x) &amp; = 9x^5 - 80x^4 + 280x^3 - 480x^2 + 405x - 124.
\end{align*}

\begin{proposition} \label{prop201}
For every integer $n \geq 688\,383$, we have
\begin{displaymath}
\frac{1}{\log p_n} \geq \frac{1}{\log n} - \frac{\log \log n}{\log^2n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n\log p_n} + \frac{1}{\log p_n} 
\sum_{k=1}^4 \frac{(-1)^{k+1}P_k(\log \log n)}{k(k+1) \log^{k+2} n}.
\end{displaymath}
\end{proposition}

\begin{proof}
We just give a sketch of the proof. For details, see \cite[Proposition 2.2]{axler2013}. We write $w = \log \log n$, $y=\log n$, and $z = \log p_n$. By 
\eqref{1.10}, the inequality $\log(1+x) \leq \sum_{k=1}^7 (-1)^{k+1} x^k/k$, which holds for every $x &gt; -1$, and the fact that $(w-1)/y + (w-2)/y^2 &gt; -1$, we 
see that
\begin{displaymath}
- y^2 + (y - w) z \leq - w^2 + (y - w) \sum_{k=1}^7 \frac{(-1)^{k+1}}{k} \left( \frac{w - 1}{y} + \frac{w - 2}{y^2} \right)^k.
\end{displaymath}
Finally, we extend the right-hand side of the last inequality to complete the proof.
\end{proof}

\begin{corollary} \label{kor202}
For every integer $n \geq 456\,914$, we have
\begin{displaymath}
\frac{1}{\log p_n} \geq \frac{1}{\log n} - \frac{\log \log n}{\log^2n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n\log p_n} 
+ \frac{P_1(\log \log n)}{2\log^3n \log p_n} - \frac{P_2(\log \log n)}{6\log^4n \log p_n}.
\end{displaymath}
\end{corollary}

\begin{proof}
See \cite[Korollar 2.6]{axler2013}.
\end{proof}

\begin{corollary} \label{kor203}
For every integer $n \geq 71$, we have
\begin{displaymath}
\frac{1}{\log p_n} \geq \frac{1}{\log n} - \frac{\log \log n}{\log^2n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n\log p_n}.
\end{displaymath}
\end{corollary}

\begin{proof}
Since the inequality
\begin{equation}
\frac{P_1(\log \log n)}{2\log n} - \frac{P_2(\log \log n)}{6\log^2 n} \geq 0 \tag{2.2} \label{2.2}
\end{equation}
holds for every $n \geq 3$, Corollary \ref{kor202} implies the validity of the required inequality for every $n \geq 456\,914$. We finish by checking the 
remaining cases with a computer.
\end{proof}

Using a similar method as in the proof of Proposition \ref{prop201}, we find the following inequality involving the reciprocal of $\log p_n$. Here, we have
\begin{align*}
P_5(x) &amp; = 3x^2-6x+5.2, \\
P_6(x) &amp; = x^3-6x^2+11.4x-4.2, \\
P_7(x) &amp; = 2x^3-7.2x^2+8.4x-4.41, \\
P_8(x) &amp; = x^3-4.2x^2+4.41x.
\end{align*}

\begin{proposition} \label{prop204}
For every integer $n \geq 2$, we have
\begin{displaymath}
\frac{1}{\log p_n} \leq \frac{1}{\log n} - \frac{\log \log n}{\log^2 n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n \log p_n} 
+ \frac{P_5(\log \log n)}{2\log^3 n \log p_n} - \sum_{k=4}^6 \frac{P_{k+2}(\log \log n)}{2\log^k n \log p_n}.
\end{displaymath}
\end{proposition}

\begin{proof}
First, we consider the case where $n \geq 33$. We write again $w = \log \log n$, $y=\log n$, and $z = \log p_n$. Notice that $\log(1+t) \geq t - t^2/2$ for 
every $t \geq 0$. If we combine the last fact with \eqref{1.11} and $(w-1)/y + (w-2.1)/y^2 \geq 0$, we obtain the inequality
\begin{displaymath}
- y^2 + (y - w)z \geq - w^2 + (y - w) \sum_{k=1}^2 \frac{(-1)^{k+1}}{k} \left( \frac{w - 1}{y} + \frac{w - 2.1}{y^2} \right)^k
\end{displaymath}
which implies the required inequality. A computer check completes the proof.
\end{proof}

Proposition \ref{prop204} implies the following both corollaries.

\begin{corollary} \label{kor205}
For every integer $n \geq 2$, we have
\begin{displaymath}
\frac{1}{\log p_n} \leq \frac{1}{\log n} - \frac{\log \log n}{\log^2 n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n \log p_n} + \frac{P_5(\log \log 
n)}{2\log^3 n \log p_n} - \sum_{k=4}^5 \frac{P_{k+2}(\log \log n)}{2\log^k n \log p_n}.
\end{displaymath}
\end{corollary}

\begin{proof}
See \cite[Korollar 2.20]{axler2013}.
\end{proof}

\begin{corollary} \label{kor206}
For every integer $n \geq 2$, we have
\begin{displaymath}
\frac{1}{\log p_n} \leq \frac{1}{\log n} - \frac{\log \log n}{\log^2 n} + \frac{(\log \log n)^2 - \log \log n + 1}{\log^2n \log p_n}
+ \frac{P_5(\log \log n)}{2\log^3 n \log p_n} - \frac{P_6(\log \log n)}{2\log^4 n \log p_n}.
\end{displaymath}
\end{corollary}

\begin{proof}
See \cite[Korollar 2.21]{axler2013}.
\end{proof}

\section{Proof of Theorem \ref{thm101}}

First, we introduce the following notation. Let the polynomials $P_1, \ldots, P_4 \in \Z[x]$ are given as in the beginning of Section 2. Let $A_0$ be a real 
number with $0.75 \leq A_0 &lt; 1$ and let $F_0 : \N \rightarrow \R$ be defined by
\begin{displaymath}
F_0(n) = \log n - A_0\log p_n.
\end{displaymath}
From \eqref{1.1}, it follows that $F_0(n)$ is nonnegative for all sufficiently large values of $n$. Let $N_0$ be a positive integer so that $F_0(n) \geq 0$ for 
every $n \geq N_0$. Furthermore, let $A_1$ be a real number with $0 &lt; A_1 \leq 458.7275$, and for $w = \log \log n$ let $F_1 : \N_{\geq 2} \rightarrow \R$ be 
given by
\begin{align*}
F_1(n) &amp; = \frac{A_1}{\log^5p_n} + \frac{(w^2 - 3.85w + 14.15)(w^2 - w + 1)}{\log^4 n\log p_n} + \frac{2.85P_1(w)}{2\log^3 n \log^2 p_n} + \frac{2.85P_1(w)}{2 
\log^4 n \log p_n} \\
&amp; \p{\q\q} + \left( \frac{13.15(w^2 - w + 1)}{\log^2 n \log^2 p_n} - \frac{70.7w}{\log^2 n \log^2 p_n}\right) \left( \frac{1}{\log n} + \frac{1}{\log p_n} 
\right) - \frac{P_2(w)}{6 \log^4 n \log p_n}.
\end{align*}
Then $F_1(n)$ is nonnegative for all sufficiently large values of $n$, and we can define $N_1$ to be a positive integer so that $F_1(n) \geq 0$ for every $n 
\geq N_1$. Further we set $A_2 = (458.7275-A_1)A_0^5$ and $A_3 = 3428.7225A_0^6$. To prove Theorem \ref{thm101}, we first use a recently obtained estimate 
\cite{ca2017} for the prime counting function $\pi(x)$ and some results from the previous section to construct a positive integer $n_0$ and an arithmetic 
function $b_0 : \N_{\geq 2} \to \R$, both depending on some parameters, with $b_0(n) \to 10.7$ as $n \to \infty$ so that
\begin{displaymath}
p_n &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n + b_0(n)}{2 \log^2 n} \right)
\end{displaymath}
for every $n \geq n_0$. In order to do this, let $a_0 : \N_{\geq 2} \rightarrow \R$ be an arithmetic function satisfying
\begin{equation}
a_0(n) \geq - (\log \log n)^2 + 6 \log \log n, \tag{3.1} \label{3.1}
\end{equation}
and let $N_2$ be a positive integer depending on the arithmetic function $a_0$ so that the inequalities
\begin{equation}
-1 &lt; \frac{\log \log n-1}{\log n} + \frac{\log \log n-2}{\log^2 n} - \frac{(\log \log n)^2 - 6\log \log n + a_0(n)}{2\log^3 n} \leq 1, \tag{3.2} \label{3.2}
\end{equation}
\begin{equation}
\frac{\log \log n - 2}{\log^2 n} - \frac{(\log \log n)^2 - 6\log \log n + a_0(n)}{2\log^3 n} \geq 0, \q \text{and} \tag{3.3} \label{3.3}
\end{equation}
\begin{equation}
p_n &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2 - 6\log \log n + a_0(n)}{2 \log^2 n} \right) \tag{3.4} 
\label{3.4}
\end{equation}
hold simultaneously for every $n \geq N_2$. Now we set
\begin{align*}
G_0(x) &amp; = \frac{2x^3 - 21x^2 + 82.2x - 98.9}{6e^{3x}} - \frac{x^4 - 14x^3 + 53.4x^2 - 100.6x + 17}{4e^{4x}} \\
&amp; \p{\q\q} + \frac{2x^5 - 10x^4 + 35x^3 - 110x^2 + 150x - 42}{10e^{5x}} - \frac{3x^4 - 44x^3 + 156x^2 - 96x + 64}{24e^{6x}},
\end{align*}
and for $w = \log \log n$ we define
\begin{align}
b_0(n) &amp; = 10.7 + \frac{2A_2}{\log^3 n} + \frac{2A_3}{\log^4 n} + \frac{a_0(n)}{\log n} \left( 1 - \frac{w- 1}{\log n} - \frac{w-2}{\log^2 n} + \frac{2w^2 - 
12w + a_0(n)}{4\log^3 n}\right) \nonumber \\
&amp; \p{\q\q} - 2\,G_0(w)\log^2 n + \frac{A_0((5.7A_0+8.7)w^2 - (32A_0 + 38)w + 147.1A_0 + 10.7)}{\log^2 n} \nonumber \\
&amp; \p{\q\q} + \frac{2 \cdot 70.7A_0^3(w^2-w+1)}{\log^4 n} + \frac{2 \cdot 70.7A_0^4(w^2-w+1)}{\log^4 n}. \tag{3.5} \label{3.5}
\end{align}
Then we obtain the following

\begin{proposition} \label{prop301}
For every integer $n \geq \max \{ N_0, N_1, N_2, 841\,424\,976 \}$, we have
\begin{displaymath}
p_n &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n + b_0(n)}{2 \log^2 n} \right).
\end{displaymath}
\end{proposition}

In order to prove this proposition, we need the following lemma. Its proof is left to the reader.

\begin{lemma} \label{lem302}
For every $x \geq 2.103$, we have
\begin{align}
0 &amp; \leq \frac{(x^2 - 3.85x + 14.15)P_1(x)}{2} - \frac{2.85 P_2(x)}{3} + \frac{P_3(x)}{12} \nonumber \\
&amp; \p{\q\q} - \frac{(x^2 - 3.85x + 14.15)P_2(x)}{6e^x} - \frac{P_4(x)}{20e^x}. \tag{3.6} \label{3.6}
\end{align}
\end{lemma}

Now we give a proof of Proposition \ref{prop301}.

\begin{proof}[Proof of Proposition \ref{prop301}]
Let $n \geq \max \{ N_0, N_1, N_2, 841\,424\,976 \}$. Using \cite[Theorem 3]{ca2017} with $x=p_n$, we see that
\begin{equation}
p_n &lt; n \left( \log p_n - 1 - \frac{1}{\log p_n} - \frac{2.85}{\log^2p_n} - \frac{13.15}{\log^3p_n} - \frac{70.7}{\log^4p_n} - \frac{458.7275}{\log^5p_n} 
- \frac{3428.7225}{\log^6p_n} \right). \tag{3.7} \label{3.7}
\end{equation}
For convenience, we write $w = \log \log n$, $y = \log n$, and $z = \log p_n$. By Corollary \ref{kor202}, we have
\begin{equation}
\frac{1}{z^2} \geq \frac{1}{yz} - \frac{w}{y^2z} + \frac{w^2-w+1}{y^2z^2} + \frac{P_1(w)}{2y^3z^2} - \frac{P_2(w)}{6y^4z^2}. \tag{3.8} \label{3.8}
\end{equation}
Again using Corollary \ref{kor202}, we get
\begin{equation}
\frac{1}{yz} \geq \Phi_1(n) = \frac{1}{y^2} - \frac{w}{y^3} + \frac{w^2-w+1}{y^3z} + \frac{P_1(w)}{2y^4z} - \frac{P_2(w)}{6y^5z}. \tag{3.9} \label{3.9}
\end{equation}
Applying \eqref{3.9} to \eqref{3.8}, we see that
\begin{equation}
\frac{1}{z^2} \geq \Phi_2(n), \tag{3.10} \label{3.10}
\end{equation}
where
\begin{displaymath}
\Phi_2(n) = \frac{1}{y^2} - \frac{w}{y^3} - \frac{w}{y^2z} + \frac{w^2-w+1}{y^3z} + \frac{w^2-w+1}{y^2z^2} + \left( \frac{P_1(w)}{2y^3z} - 
\frac{P_2(w)}{6y^4z} \right) \left( \frac{1}{y} + \frac{1}{z} \right).
\end{displaymath}
Now \eqref{2.2} implies that
\begin{equation}
\frac{1}{z^2} \geq \Phi_3(n) = \frac{1}{y^2} - \frac{w}{y^3} - \frac{w}{y^2z} + \frac{w^2-w+1}{y^3z} + \frac{w^2-w+1}{y^2z^2}. \tag{3.11} \label{3.11}
\end{equation}
We assumed $n \geq N_0$. Hence $F_0(n) \geq 0$, which is equivalent to
\begin{equation}
\frac{A_0}{y} \leq \frac{1}{z}. \tag{3.12} \label{3.12}
\end{equation}
From \eqref{3.12} and the fact that $2.85x^2 - 16x + 73.55 \geq 0$ for every $x \geq 0$, it follows
\begin{equation}
\frac{2.85w^2-16w+73.55}{z^2} \geq \frac{A_0(5.7w^2 - 32w + 147.1)}{2yz}. \tag{3.13} \label{3.13}
\end{equation}
Let $f(x) = (5.7A_0+8.7)x^2 - (32A_0 + 38)x + 147.1A_0 + 10.7$. Since $0.75 \leq A_0 &lt; 1$, we get $f(x) \geq 12.975x^2 - 70x + 121.025 \geq 0$ for every $x 
\geq 0$. Using \eqref{3.12} and \eqref{3.13}, we get
\begin{align}
&amp; \frac{2.85w^2 - 16w + 73.55}{z^2} + \frac{8.7w^2 - 38w + 10.7}{2yz} \geq \frac{A_0f(w)}{2y^2}. \tag{3.14} \label{3.14}
\end{align}
We recall that $A_2 = (458.7275-A_1)A_0^5$ and $A_3 = 3428.7225A_0^6$. Hence \eqref{3.12} implies that
\begin{equation}
\frac{A_2}{y^5} + \frac{A_3}{y^6} + \frac{70.7A_0^3}{y^6} + \frac{70.7A_0^4}{y^6} \leq \frac{458.7275 - A_1}{z^5} + \frac{3428.7225}{z^6} + \frac{70.7}{y^3z^3} 
+ \frac{70.7}{y^2z^4}. \tag{3.15} \label{3.15}
\end{equation}
Now we apply \eqref{3.14} and \eqref{3.15} to \eqref{3.5} and see that
\begin{align}
\frac{10.7 - b_0(n)}{2y^2} &amp; + \frac{2.85(w^2 - w + 1)}{y^2z^2} - \frac{13.15w}{y^2z^2} + \frac{70.7}{y^2z^2} + \frac{8.7w^2 - 38w + 10.7}{2y^3z} \nonumber \\
&amp; \p{\q\q} + \frac{458.7275-A_1}{z^5} + \frac{3428.7225}{z^6} + \frac{70.7(w^2-w+1)}{y^2z^3} \left( \frac{1}{y} + \frac{1}{z} \right) 
\nonumber \\
&amp; \geq G_0(w) - \frac{a_0(n)}{2y^3} \left( 1 - \frac{w- 1}{y} - \frac{w-2}{y^2} + \frac{2w^2 - 12w + a_0(n)}{4y^3}\right). \tag{3.16} \label{3.16}
\end{align}
The inequality \eqref{2.2} tells us that
\begin{equation}
\frac{13.15}{z} \left( \frac{P_1(w)}{2y^3z} - \frac{P_2(w)}{6y^4z} \right)\left( \frac{1}{y} + \frac{1}{z} \right) \geq 0. \tag{3.17} \label{3.17}
\end{equation}
Adding the left-hand side of \eqref{3.17} and the right-hand side of \eqref{3.6} with $x = w$ to the left-hand side of \eqref{3.16}, we get
\begin{align*}
\frac{5.35}{y^2} &amp;- \frac{b_0(n)}{2y^2} + \frac{2.85(w^2 - w + 1)}{y^2z^2} - \frac{13.15w}{y^2z^2} + \frac{70.7}{y^2z^2} + \frac{8.7w^2 - 38w + 10.7}{2y^3z} + 
\frac{458.7275-A_1}{z^5} \\
&amp; \p{\q\q} + \frac{3428.7225}{z^6} + \frac{70.7(w^2 - w + 1)}{y^2z^3} \left( \frac{1}{y} + \frac{1}{z} \right) + \frac{13.15}{z} \left( 
\frac{P_1(w)}{2y^3z} - \frac{P_2(w)}{6y^4z} \right)\left( \frac{1}{y} + \frac{1}{z} \right) \\
&amp; \p{\q\q} - \frac{2.85 P_2(w)}{6 y^5z} - \frac{2.85P_2(w)}{6y^4z^2} + \frac{(w^2-3.85w+14.15)P_1(w)}{2 y^5z} + \frac{P_3(w)}{12y^5z} - \frac{P_4(w)}{20y^6z} \\
&amp; \p{\q\q} - \frac{(w^2-3.85w+14.15)P_2(w)}{6y^6z} \\
&amp; \geq G_0(w) - \frac{a_0(n)}{2y^3} \left( 1 - \frac{w- 1}{y} - \frac{w-2}{y^2} + \frac{2w^2 - 12w + a_0(n)}{4y^3}\right).
\end{align*}
Since $n \geq N_1$, we have $F_1(n) \geq 0$. Now we add $F_1(n)$ to the left-hand side of the last inequality, use the identity $8.7w^2 - 38w + 10.7 = P_1(w) + 
2\cdot 2.85(w^2-w+1) - 2 \cdot 13.15w$, and collect all terms containing the number $70.7$ and the term $w^2-3.85w+14.15$, respectively, to get
\begin{align*}
\frac{5.35}{y^2} &amp;- \frac{b_0(n)}{2y^2} + \frac{2.85(w^2 - w + 1)}{y^2z^2} - \frac{13.15w}{y^2z^2} + \frac{70.7}{z^2} \cdot \Phi_3(n) + \frac{458.7275}{z^5} + 
\frac{3428.7225}{z^6} \\
&amp; \p{\q\q} + \frac{2.85(w^2 - w + 1)}{y^3z} - \frac{13.15w}{y^3z} + \left( 2.85 + \frac{13.15}{z} \right) \left( \frac{P_1(w)}{2y^3z} - \frac{P_2(w)}{6y^4z} 
\right)\left( \frac{1}{y} + \frac{1}{z} \right) \\
&amp; \p{\q\q} + \frac{w^2-3.85w+14.15}{y} \cdot \Phi_1(n) + \frac{P_1(w)}{2y^3z} - \frac{P_2(w)}{6 y^4z} + \frac{P_3(w)}{12y^5z} - \frac{P_4(w)}{20y^6z} \\
&amp; \p{\q\q} + \frac{13.15(w^2 - w + 1)}{y^2z^2}\left( \frac{1}{y} + \frac{1}{z} \right) - \frac{2.85w}{y^3} \\
&amp; \geq \widetilde{G_0}(w) - \frac{a_0(n)}{2y^3} \left( 1 - \frac{w- 1}{y} - \frac{w-2}{y^2} + \frac{2w^2 - 12w + a_0(n)}{4y^3}\right),
\end{align*}
where $\Phi_1(n)$ and $\Phi_3(n)$ are given as in \eqref{3.9} and \eqref{3.11}, respectively, and 
\begin{displaymath}
\widetilde{G_0}(x) = G_0(x) + \frac{x^2-3.85x + 14.15}{e^{3x}}- \frac{x^3 - 3.85x^2 + 14.15x}{e^{4x}} - \frac{2.85x}{e^{3x}}.
\end{displaymath}
Now we use \eqref{3.9} and \eqref{3.11} and collect all terms containing the numbers $2.85$ and $13.15$ to see that
\begin{align*}
\frac{2.5}{y^2} &amp;- \frac{b_0(n)}{2y^2} + \left( 2.85 + \frac{13.15}{z} \right) \Phi_2(n) + \frac{70.7}{z^4} + \frac{458.7275}{z^5} + \frac{3428.7225}{z^6} 
+ \frac{w^2-w+1}{y^2z} \\
&amp; \p{\q\q} + \frac{P_1(w)}{2y^3z} - \frac{P_2(w)}{6 y^4z} + \frac{P_3(w)}{12y^5z} - \frac{P_4(w)}{20y^6z} \\
&amp; \geq \widetilde{G_0}(w) - \frac{a_0(n)}{2y^3} \left( 1 - \frac{w- 1}{y} - \frac{w-2}{y^2} + \frac{2w^2 - 12w + a_0(n)}{4y^3}\right).
\end{align*}
Applying \eqref{3.10} and Proposition \ref{prop201}, we get
\begin{align*}
\frac{2.5}{y^2} &amp; - \frac{b_0(n)}{2y^2} + \frac{2.85}{z^2} + \frac{13.15}{z^3} + \frac{70.7}{z^4} + \frac{458.7275}{z^5} + \frac{3428.7225}{z^6} - \frac{1}{y} 
+ \frac{w}{y^2} + \frac{1}{z} \\
&amp; \geq \widetilde{G_0}(w) - \frac{a_0(n)}{2y^3} \left( 1 - \frac{w- 1}{y} - \frac{w-2}{y^2} + \frac{2w^2 - 12w + a_0(n)}{4y^3}\right).
\end{align*}
A straightforward calculation shows that the last inequality is equivalent to
\begin{align*}
- \frac{1}{y} &amp; - \frac{w^2 - 4w - (4-b_0(n))}{2y^2} + \frac{1}{z} + \frac{2.85}{z^2} + \frac{13.15}{z^3} + \frac{70.7}{z^4} + \frac{458.7275}{z^5} + 
\frac{3428.7225}{z^6} \nonumber\\
&amp; \geq - \frac{w^2 - 6w+a_0(n)}{2y^3} - \frac{1}{2} \left( \frac{w - 1}{y} + \frac{w - 2}{y^2} - \frac{w^2 - 6w + a_0(n)}{2y^3} \right)^2 \nonumber\\
&amp; \p{\q\q}  + \frac{1}{3} \left(
\frac{w - 1}{y} + \frac{w - 2}{y^2} \right)^3 - \frac{1}{4} \left( \frac{w - 1}{y} \right)^4 + \frac{1}{5} \left(  \frac{w - 1}{y} \right)^5.
\end{align*}
We add $(w - 1)/y + (w - 2)/y^2$ to both sides of this inequality. Since $\log(1+x) \leq \sum_{k=1}^5 (-1)^{k+1}x/k$ for every $x &gt; -1$, $g(x) = x^3/3$ is 
increasing, and $h(x)= - x^4/4 + x^5/5$ is decreasing on the interval $[0,1]$, we can use \eqref{3.1}--\eqref{3.3} to get
\begin{align*}
y &amp; + w - 1 + \frac{w - 2}{y} - \frac{w^2 - 6w + b_0(n)}{2y^2} + \frac{1}{z} + \frac{2.85}{z^2} + \frac{13.15}{z^3} + \frac{70.7}{z^4} + \frac{458.7275}{z^5} + 
\frac{3428.7225}{z^6} \\
&amp; \geq y + w - 1 + \log \left( 1 + \frac{w - 1}{y} + \frac{w - 2}{y^2} - \frac{w^2 - 6w + a_0(n)}{2y^3} \right).
\end{align*}
Finally, we use \eqref{3.4} and \eqref{3.7} to arrive at the desired result.
\end{proof}

Next we use Proposition \ref{prop301} and the following both lemmata to prove Theorem \ref{thm101}. In the first lemma we determine a suitable value of $N_0$ 
for $A_0 = 0.87$.

\begin{lemma} \label{lem303}
For every integer $n \geq 1\,338\,564\,587$, we have
\begin{displaymath}
\log n \geq 0.87 \log p_n.
\end{displaymath}
\end{lemma}

\begin{proof}
We set
\begin{displaymath}
f(x) = e^x - 0.87 \left( e^x + x + \log \left( 1 + \frac{x-1}{e^x} + \frac{x-2}{e^{2x}} \right) \right).
\end{displaymath}
Since $f'(x) \geq 0$ for every $x \geq 2.5$ and $f(3.046) \geq 0.00137$, we see that $f(x) \geq 0$ for every $x \geq 3.046$. Substituting $x = \log \log n$ in 
$f(x)$ and using \eqref{1.10}, we see that the desired inequality holds for every $n \geq \exp(\exp(3.046))$. We check the remaining cases with a computer.
\end{proof}

Now we use Lemma \ref{lem303} to find a suitable value of $N_1$ for $A_1 = 155.32$. 

\begin{lemma} \label{lem304}
Let $A_1 = 155.32$. Then $F_1(n) \geq 0$ for every $n \geq 100\,720\,878$.
\end{lemma}

\begin{proof}
First, let $n \geq \exp(\exp(3.05))$. We have
\begin{align*}
F_1(n) &amp; = \frac{155.32}{z^5} + \frac{f(w)}{6 y^4z} + \frac{34.85w^2 - 184.8w+40.55}{2y^3z^2} + \frac{13.15w^2 - 83.85w+13.15}{y^2z^3}.
\end{align*}
where $f(x) = 6x^4 - 34.1x^3 + 163.65x^2 - 198.3x + 141.65$. Since $f(x) \geq 0$ for every $x \geq 3.05$, it suffices to show that
\begin{equation}
\frac{155.32}{z^5} + \frac{6w^4 - 34.1w^3 + 268.2w^2 - 752.7w + 263.3}{6y^3z^2} + \frac{13.15w^2 - 83.85w + 13.15}{y^2 z^3} \geq 0. \tag{3.18} \label{3.18}
\end{equation}
In order to do this, we set
\begin{align*}
g(x) &amp; = (6x^4 - 34.1x^3 + 268.2x^2 - 752.7x + 263.3)(e^x+x) \\
&amp; \p{\q\q} + 6e^x(13.15x^2-83.85x + 13.15 + 155.32 \cdot 0.87^2).
\end{align*}
It is easy to see that $h_1(x) = 6x^4 - 10.1x^3 + 244.8x^2 - 561.6x - 208.229752 \geq 0$ for every $x \geq 2.6$ and that $h_2(x) = 30x^4 - 136.4x^3 + 804.6x^2 
- 1505.4x + 263.3 \geq 0$ for every $x \geq 2.2$. Hence $g'(x) = h_1(x)e^x + h_2(x) \geq 0$ for every $x \geq 2.6$. We also have $g(3.05) \geq 0.9$. Therefore, 
$g(x) \geq 0$ for every $x \geq 3.05$. Since $6x^4 - 34.1x^3 + 268.2x^2 - 752.7x + 263.3 \geq 0$ for every $x \geq 3.05$, we can use \eqref{1.3} to get 
$g(w)/(6y^3z^3) \geq 0$. Now we apply Lemma \ref{lem303} to obtain \eqref{3.18}. We finish by direct computation.
\end{proof}

Finally, we give a proof of Theorem \ref{thm101}.

\begin{proof}[Proof of Theorem \ref{thm101}]
For convenience, we write $w= \log \log n$ and $y=\log n$. Setting $A_0 = 0.87$ and $A_1 = 155.32$, we use Lemma \ref{lem303} and Lemma \ref{lem304} to get 
$N_0 =1\,338\,564\,587$ and $N_1 = 100\,720\,878$, respectively. The proof of this theorem goes in two steps.

\textit{Step 1}. We set $a_0(n) = - w^2+6w$. Then $N_2 = 688\,383$ is a suitable choice for $N_2$. By \eqref{3.5}, we get
\begin{equation}
b_0(n) \geq 10.7 + g(n), \tag{3.19} \label{3.19}
\end{equation}
where
\begin{align*}
g(n) &amp; = - \frac{2w^3-18w^2+64.2w-98.9}{3y} + \frac{w^4 - 12w^3 + 63.16w^2 - 203.17w + 258.29}{2y^2} \\
&amp; \p{\q\q} - \frac{2w^5 - 10w^4 + 30w^3 - 70w^2 + 90w - 1554.24}{5y^3} \\
&amp; \p{\q\q} - \frac{8w^3 - 2137.44w^2 + 2185.45w - 37836.25}{12y^4}.
\end{align*}
We define
\begin{align*}
g_1(x,t) &amp; = 3.54e^{4x} + 20(18x^2 + 98.9)e^{3x} -20(2t^3 + 64.2t)e^{3t} \\
&amp; \p{\q\q} + 30(x^4 + 63.16x^2 + 258.29)e^{2x} - 30(12t^3 + 203.17t)e^{2t} \\
&amp; \p{\q\q} + 12(10x^4 + 70x^2 + 1554.24)e^x - 12(2t^5 + 30t^3 + 90t)e^t \\
&amp; \p{\q\q} + 5(2137.44x^2 + 37836.25) - 5(8t^3 + 2185.45t).
\end{align*}
If $t_0 \leq x \leq t_1$, then $g_1(x,x) \geq g_1(t_0,t_1)$. We check with a computer that $g_1(i \cdot 10^{-5}, (i+1) \cdot 10^{-5}) \geq 0$ for every 
integer $i$ with $0 \leq i \leq 699\,999$. Therefore,
\begin{equation}
g(n) + 0.059 = \frac{g_1(w,w)}{60y^4} \geq 0 \q\q (0 \leq w \leq 7). \tag{3.20} \label{3.20}
\end{equation}
Next we prove that $g_1(x,x) \geq 0$ for every $x \geq 7$. For this purpose, let $W_1(x) = 3.54e^x -20(2x^3 - 18x^2 + 64.2x - 98.9)$. It is easy to show that 
$W_1(x) \geq 792$ for every $x \geq 7$. Hence we get
\begin{align*}
g_1(x,x) &amp; \geq (792e^x + 30(x^4 - 12x^3 + 63.16x^2 - 203.17x + 258.29))e^{2x} \\
&amp; \p{\q\q} - 12(2x^5 - 10x^4 + 30x^3 - 70x^2 + 90x - 1554.24)e^x \\
&amp; \p{\q\q} - 5(8x^3 - 2137.44x^2 + 2185.45x - 37836.25).
\end{align*}
Since $792e^t + 30(t^4 - 12t^3 + 63.16t^2 - 203.17t + 258.29) \geq 875\,011$ for every $t \geq 7$, we obtain $g(n) + 0.059 = g_1(w,w)/(60y^4) \geq 0$ for $w 
\geq 7$. Combined with \eqref{3.20}, it gives that $g(n) \geq -0.059$ for every $n \geq 3$. Applying this to \eqref{3.19}, we get $b_0(n) \geq 10.641$ for every 
$n \geq 3$. Hence, by Proposition \ref{prop301}, we get
\begin{displaymath}
p_n &lt; n \left( y + w - 1 + \frac{w - 2}{y} - \frac{w^2-6w + 10.641}{2y^2} \right)
\end{displaymath}
for every $n \geq 1\,338\,564\,587$. For every integer $n$ such that $39\,529\,802 \leq n \leq 1\,338\,564\,586$ we check the last inequality 
with a computer.

\textit{Step 2}. We set $a_0(n) = 10.641$. Using the first step, we can choose $N_2 = 39\,529\,802$. By \eqref{3.5}, we have
\begin{equation}
b_0(n) \geq 10.7 + h(n), \tag{3.21} \label{3.21}
\end{equation}
where $h(n)$ is given by
\begin{align*}
h(n) &amp; = - \frac{2w^3-21w^2+82.2w-130.823}{3y} + \frac{w^4 - 14w^3 + 77.16w^2 - 236.45w + 279.57}{2y^2} \\
&amp; \p{\q\q} - \frac{2w^5-10w^4+35w^3-110w^2+203.205w-1660.65}{5y^3} \\
&amp; \p{\q\q} + \frac{3w^4 - 44w^3 + 2309.28w^2 - 2568.52w + 38175.947}{12y^4}.
\end{align*}
We set
\begin{align*}
h_1(x,t) &amp; = 1.98e^{4x} + 20(21x^2+130.823)e^{3x} - 20(2t^3+82.2t)e^{3t} \\
&amp; \p{\q\q} + 30(x^4 + 77.16x^2 + 279.57)e^{2x} - 30(14t^3 + 236.45t)e^{2t} \\
&amp; \p{\q\q} + 12(10x^4+110x^2+1660.65)e^x - 12(2t^5+35t^3+203.205t)e^t \\
&amp; \p{\q\q} + 5(3x^4 + 2309.28x^2 + 38175.947) - 5(44t^3 + 2568.52t).
\end{align*}
Clearly, $h_1(x,x) \geq h_1(t_0,t_1)$ for every $x$ such that $t_0 \leq x \leq t_1$. We use a computer to verify that $h_1(i \cdot 10^{-6}, (i+1) \cdot 
10^{-6}) 
\geq 0$ for every integer $i$ with $0 \leq i \leq 7\,999\,999$. Therefore,
\begin{equation}
h(n) + 0.033 = \frac{h_1(w,w)}{60y^4} \geq 0 \q\q (0 \leq w \leq 8). \tag{3.22} \label{3.22}
\end{equation}
We next show that $h_1(x,x) \geq 0$ for every $x \geq 8$. Since $1.98e^t -20(2t^3 - 21t^2 + 82.2t - 130.823) \geq 1766$ for every $t \geq 8$, we have
\begin{align*}
h_1(x,x) &amp; \geq 1766e^{3x} + 30(x^4 - 14x^3 + 77.16x^2 - 236.45x + 279.57)e^{2x} \\
&amp; \p{\q\q} - 12(2x^5 - 10x^4 + 35x^3 - 110x^2 + 203.205x - 1660.65)e^x \\
&amp; \p{\q\q} + 5(3x^4 - 44x^3 + 2309.28x^2 - 2568.52x + 38175.947).
\end{align*}
Note that $1766e^t + 30(t^4 - 14t^3 + 77.16t^2 - 236.45t + 279.57) \geq 5\,271\,998$ for every $t \geq 8$. Hence $h(n) + 0.033 = h_1(w,w)/(60y^4) \geq 0$ for 
$w \geq 8$. Combined with \eqref{3.22} and \eqref{3.21}, this gives $b_0(n) \geq 10.667$ for every $n \geq 3$. Applying this to Proposition \ref{prop301}, we 
complete the proof of the required inequality for every $n \geq 1\,338\,564\,587$. We verify the remaining cases with a computer.
\end{proof}

Denoting the right-hand side of \eqref{1.10} by $D_{\rm up}(n)$ and the right-hand side \eqref{1.12} by $A_{up}(n)$, we use \seqnum{A006988} to compare the error 
term of the approximation from Theorem \ref{thm101} with Dusart's approximation from \eqref{1.10} for the $10^n$th prime number:
\begin{center}
\begin{tabular}{|c||r|r|r|}
\hline
$n$\rule{0mm}{4mm} &amp; $p_n$ &amp; $\lceil D_{\rm up}(n) -p_n\rceil $ &amp; $\lceil  A_{up}(n)-p_n \rceil$ \\ \hline
$ 10^{10} $\rule{0mm}{4mm} &amp; $252\,097\,800\,623$ &amp; $20\,510\,784$ &amp; $4\,613\,984$ \\\hline
$ 10^{11} $\rule{0mm}{4mm} &amp; $2\,760\,727\,302\,517$ &amp; $172\,884\,400$ &amp; $38\,768\,198$ \\\hline
$ 10^{12} $\rule{0mm}{4mm} &amp; $29\,996\,224\,275\,833$ &amp; $1\,469\,932\,710$ &amp; $311\,593\,524$ \\\hline
$ 10^{13} $\rule{0mm}{4mm} &amp; $323\,780\,508\,946\,331$ &amp; $12\,732\,767\,836$ &amp; $2\,542\,231\,421$ \\\hline
$ 10^{14} $\rule{0mm}{4mm} &amp; $3\,475\,385\,758\,524\,527$ &amp; $112\,026\,014\,682$ &amp; $21\,049\,069\,521$ \\\hline
$ 10^{15} $\rule{0mm}{4mm} &amp; $37\,124\,508\,045\,065\,437$ &amp; $998\,861\,791\,991$ &amp; $176\,995\,293\,694$ \\\hline
$ 10^{16} $\rule{0mm}{4mm} &amp; $394\,906\,913\,903\,735\,329$ &amp; $9\,004\,342\,407\,404$ &amp; $1\,507\,803\,850\,451$ \\\hline
$ 10^{17} $\rule{0mm}{4mm} &amp; $4\,185\,296\,581\,467\,695\,669$ &amp; $81\,924\,060\,077\,026$ &amp; $12\,998\,658\,322\,559$ \\\hline
$ 10^{18} $\rule{0mm}{4mm} &amp; $44\,211\,790\,234\,832\,169\,331$ &amp; $751\,154\,982\,343\,786$ &amp; $113\,204\,602\,033\,556$ \\\hline
$ 10^{19} $\rule{0mm}{4mm} &amp; $465\,675\,465\,116\,607\,065\,549$ &amp; $6\,932\,757\,377\,044\,654$ &amp; $994\,838\,584\,902\,026$ \\\hline
$ 10^{20} $\rule{0mm}{4mm} &amp; $4\,892\,055\,594\,575\,155\,744\,537$ &amp; $64\,346\,895\,915\,006\,577$ &amp; $8\,812\,315\,669\,274\,243$ \\ \hline
\end{tabular}
\end{center}

\section{Proof of Theorem \ref{thm103}}

In order to do prove Theorem \ref{thm103}, we introduce the \emph{logarithmic integral} $\text{li}(x)$ which is defined for every real $x \geq 0$ as
\begin{displaymath}
\text{li}(x) = \int_0^x \frac{\text{d}t}{\log t} = \lim_{\e \to 0+} \left \{ \int_0^{1-\e}{\frac{\text{d}t}{\log t}} + \int_{1+\e}^x{\frac{\text{d}t}{\log 
t}} \right \}.
\end{displaymath}

\begin{proof}[Proof of Theorem \ref{thm103}]
Let $x_0 = 3\,273\,361\,096$. First, we verify the required inequality for every integer $n$ with $x_0 \leq n \leq \pi(10^{19})$. For $x &gt; 1$, the logarithmic 
integral $\tl(x)$ is increasing with $\tl((1, \infty)) = \R$. Thus, we can define the inverse function $\tl^{-1} : \R \to (1, \infty)$ by
\begin{equation}
\tl(\tl^{-1}(x)) = x. \tag{4.1} \label{4.1}
\end{equation}
Further, let
\begin{displaymath}
f(x) = x - \text{li} \left( x \left( \log x + \log \log x - 1 + \frac{\log \log x - 2}{\log x} - \frac{(\log \log x)^2 - 6\log \log x + 11.25}{2 \log^2 x} 
\right) \right).
\end{displaymath}
We show that $f(x) &gt; 0$ for every $x \geq x_0$. We have $f(x_0) &gt; 0.000001$. So it suffices to show that $f'(x) \geq 0$ for every $x \geq x_0$. Setting
\begin{displaymath}
g_1(a,b) = \log \left( 1 + \frac{\log a - 1}{a} + \frac{\log a - 2}{a^2} - \frac{\log^2b - 6\log b + 11.25}{2b^3} \right)
\end{displaymath}
and $g(z) = g_1(z,z)$, we see that $(z + \log z + g(z))f'(e^z) = h(z)$, where
\begin{displaymath}
h(z) = g(z) - \frac{\log z - 1}{z} + \frac{\log^2z - 4\log z + 5.25}{2 z^2} - \frac{\log^2z - 7\log z + 14.25}{z^3}.
\end{displaymath}
Since $z + \log z + g(z) &gt; 0$ for every $z \geq 2.1$, it suffices to verify that $h(z) \geq 0$ for every $z \geq \log x_0$. We have $h(\log x_0) \geq 0.000026$ 
and
\begin{align}
(-4)z^7e^{g(z)}h'(z) &amp; = z^4 + (4 \log^3 z - 46 \log^2 z + 197 \log z - 323.5)z^3 \nonumber \\
&amp; \p{\q\q} + (-6\log^3 z + 60\log^2 z - 175.5 \log z + 90)z^2 \nonumber \\
&amp; \p{\q\q} + (-2 \log^4 z + 10 \log^3 z + 19 \log^2 z - 183.5 \log z + 234.876)z \nonumber \\
&amp; \p{\q\q} + 6 \log^4 z - 82 \log^3 z + 443 \log^2 z - 1114.5 \log z + 1119.375. \tag{4.2} \label{4.2}
\end{align}
In order to show that $h'(z) &gt; 0$ for every $z \in J = [\log x_0, 29.8]$, it suffices to show that the right-hand side of \eqref{4.2} is negative. Since $z + 
4 \log^3 z - 46 \log^2 z + 197 \log z - 325.5 &lt; 1.43$ for every $z \in J$, we get
\begin{align*}
(-4)z^7e^{g(z)}h'(z) &amp; &lt; 1.43z^3 + (-6\log^3 z + 60\log^2 z - 175.5 \log z + 90)z^2 \\
&amp; \p{\q\q} + (-2 \log^4 z + 10 \log^3 z + 19 \log^2 z - 183.5 \log z + 234.876)z \\
&amp; \p{\q\q} + 6 \log^4 z - 82 \log^3 z + 443 \log^2 z - 1114.5 \log z + 1119.375.
\end{align*}
Notice that $1.43z -6 \log^3 z + 60\log^2 z - 175.5 \log z + 90 \leq -0.444$ for every $z \in J$. Hence
\begin{align*}
(-4)z^7e^{g(z)}h'(z) &amp; &lt; -0.444z^2 + (-2 \log^4 z + 10 \log^3 z + 19 \log^2 z - 183.5 \log z + 234.876)z \\
&amp; \p{\q\q} + 6 \log^4 z - 82 \log^3 z + 443 \log^2 z - 1114.5 \log z + 1119.375.
\end{align*}
We have $-0.444z - 2 \log^4 z + 10 \log^3 z + 19 \log^2 z - 183.5 \log z + 234.876 \leq - 47.701$ for every $z \in J$. Hence $(-4)z^7e^{g(z)}h'(z) &lt; 0$ for 
every $z \in J$ which yields that $h'(z) &gt; 0$ for every $z \in J$. Combined with $h(\log x_0) &gt; 0$, it turns out that $h(z) &gt; 0$ for every $z \in [\log x_0 , 
29.8]$. Similar, we get $h'(z) &lt; 0$ for every $z \geq 29.88$. Together with $\lim_{z \to \infty} h(z) = 0$, we see that $h(z) \geq 0$ for every $z \geq 29.88$. 
It remains to consider the case where $z \in (29.8, 29.88)$. If $a \leq z \leq b$, then
\begin{displaymath}
h(z) \geq h_1(a,b) = g_1(a,b) - \frac{\log b - 1}{b} + \frac{\log^2a - 4\log a + 5.25}{2a^2} - \frac{\log^2b - 7\log b + 14.25}{b^3}.
\end{displaymath}
Now we check with a computer that $h_1(29.8, 29.88) &gt; 0$. Hence $f(x) &gt; 0$ for every $x \geq x_0$. Since $\tl(x)$ is increasing for $x &gt; 1$, we can use 
\eqref{4.1} to get
\begin{displaymath}
x \left( \log x + \log \log x - 1 + \frac{\log \log x - 2}{\log x} - \frac{(\log \log x)^2 - 6\log \log x + 11.25}{2 \log^2 x} \right) &lt; \text{li}^{-1}(x)
\end{displaymath}
for every $x \geq x_0$. Applying \cite[Lemma 7]{pomerance} to the last inequality, we see that the desired inequality holds for every integer $n$ satisfying 
$3\,273\,361\,096 \leq n \leq \pi(10^{19})$. For every integer $n$ such that $2 \leq n &lt; 3\,273\,361\,096$ we check the desired inequality with a computer.
\end{proof}


\section{Proof of Theorem \ref{thm104}}

Compared with the proof of Theorem \ref{thm103}, the proof of Theorem \ref{thm104} is rather technical and we need to introduce some notation. First, let
\begin{align*}
P_9(x) &amp; = P_5(x) + 2 \cdot 3.15(x^2-x+1), \\
P_{10}(x) &amp; = (x^2-x+1)P_9(x) + (x^2-x+1)P_5(x) - 3.15P_6(x) - P_7(x) + 12.85P_5(x), \\
P_{11}(x) &amp; = 3.15P_7(x) + 12.85P_6(x), \\
P_{12}(x) &amp; = 2(x^2-x+1)P_6(x) - P_5(x)P_9(x),
\end{align*}
where the polynomials $P_5$, $P_6$, $P_7$, and $P_8$ were defined as in Section 2. Let $B_1, \ldots, B_{10}$ be real positive constants satisfying
\begin{equation} 
B_6 + B_7 + B_8 + B_9 + B_{10} \leq 3.15. \tag{5.1} \label{5.1}
\end{equation}
Writing $w = \log \log n$, $y = \log n$, and $z = \log p_n$, we define $H_i : \N_{\geq 2} \rightarrow \R$, where $1 \leq i \leq 10$, by
\begin{itemize}
\item $\displaystyle H_1(n) = \frac{B_1w}{y^3z} - \frac{P_{10}(w)}{2y^5z} + \frac{P_{11}(w)}{2y^5z^2} + \frac{P_{12}(w)}{4y^6z} + \frac{12.85P_6(w)}{2y^4z^3}$,
\item $\displaystyle H_2(n) = \frac{B_2w}{y^3z} + \frac{12.85w}{y^2z^2} - \frac{71.3}{z^4}$,
\item $\displaystyle H_3(n) = \frac{B_3w}{y^3z} - \frac{3.15P_5(w)}{2y^3z^2} - \frac{12.85(w^2 - w+1)}{y^3z^2}$,
\item $\displaystyle H_4(n) = \frac{B_4w}{y^3z} + \frac{3.15P_6(w) - 12.85P_5(w)}{2y^4z^2}$,
\item $\displaystyle H_5(n) = \frac{B_5w}{y^3z} + \frac{P_6(w) - 3.15P_5(w)}{2y^4z} - \frac{12.85(w^2 - w +1)}{y^4z} - \frac{(w^2-w+1)^2}{y^4z}$,
\item $\displaystyle H_6(n) = \frac{B_6w}{y^2z} + \frac{(12.85-B_1-B_2-B_3-B_4-B_5)w}{y^3z} - \frac{3.15(w^2 - w +1)}{y^2z^2}$,
\item $\displaystyle H_7(n) = \frac{B_7w}{y^2z} - \frac{12.85 P_5(w)}{2y^3z^3}$,
\item $\displaystyle H_8(n) = \frac{B_8w}{y^2z} - \frac{12.85(w^2 - w +1)}{y^2z^3}$,
\item $\displaystyle H_9(n) = \frac{B_9w}{y^2z} - \frac{463.2275}{z^5}$,
\item $\displaystyle H_{10}(n) = \frac{B_{10}w}{y^2z} - \frac{4585}{z^6}$.
\end{itemize}
Then $H_i(n)$, $1 \leq i \leq 10$, is nonnegative for all sufficiently large values of $n$. Let $K_1$ be a positive integer so that $H_i(n) \geq 0$, $1 \leq i 
\leq 10$, for every $n \geq K_1$. Let $a_1 : \N_{\geq 2} \rightarrow \R$ be an arithmetic function and let $K_2$ be a positive integer, which 
depends on $a_1$, so that the inequalities
\begin{equation}
a_1(n) &gt; - (\log \log n)^2 + 6\log \log n, \tag{5.2} \label{5.2}
\end{equation}
\begin{equation}
0 \leq \frac{\log \log n-1}{\log n} + \frac{\log \log n - 2}{\log^2 n} - \frac{(\log \log n)^2 - 6\log \log n+a_1(n)}{2\log^3 n} \leq 1, \q \text{and}  
\tag{5.3} \label{5.3}
\end{equation}
\begin{equation}
p_n &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n+a_1(n)}{2\log^2 n} \right) \tag{5.4} \label{5.4}
\end{equation}
hold simultaneously for every $n \geq K_2$. Furthermore, we define the function $G_1 : \R \rightarrow \R$ by
\begin{align*}
G_1(x) &amp; = \frac{3.15x}{e^{3x}} - \frac{12.85}{e^{3x}} + \frac{12.85x}{e^{4x}} - \frac{x^2 - x + 1}{e^{3x}} + \frac{(x^2-x+1)x}{e^{4x}} - 
\frac{P_9(x)}{2e^{4x}} + \frac{P_9(x)x}{2e^{5x}} \\
&amp; \p{\q\q} + \frac{(x-1)^2}{2e^{2x}} - \frac{x^2-6x}{2e^{3x}} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( \frac{x-1}{e^x} + 
\frac{x-2}{e^{2x}} \right)^k  + \frac{(x-2)^4}{4e^{8x}}.
\end{align*}
In order to prove Theorem \ref{thm104}, we set
\begin{align}
b_1(n) &amp; = 11.3 - 2\,G_1(\log \log n) \log^2 n + \frac{a_1(n)}{\log n} \nonumber \\
&amp; \p{\q\q} - \frac{2A_0(3.15-(B_6 + B_7 + B_8 + B_9 + B_{10}))\log \log n}{\log n} \tag{5.5} \label{5.5}
\end{align}
and prove the following proposition.

\begin{proposition} \label{prop401}
For every integer $n \geq \max \{N_0, K_1, K_2, 3520\}$,  we have
\begin{displaymath}
p_n &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2-6\log \log n + b_1(n)}{2 \log^2 n} \right).
\end{displaymath}
\end{proposition}

The following lemma is helpful for the proof of Proposition \ref{prop401}. The proof is left to the reader.

\begin{lemma} \label{lem402}
Let $w = \log \log n$. For every integer $n \geq 6$, we have
\begin{displaymath}
\frac{12.85P_6(w)}{2\log^6 n \log p_n} + \frac{3.15P_7(w)}{2\log^6 n \log p_n} + \frac{P_8(w)}{2\log^6 n \log p_n} \geq 0,
\end{displaymath}
and for every integer $n \geq 17$, we have
\begin{displaymath}
\frac{P_6(w)P_9(w)}{4\log^7n\log p_n} + \frac{12.85P_7(w)}{2\log^7n\log p_n} + \frac{3.15P_8(w)}{2\log^7n\log p_n} + 
\frac{3.15P_8(w)}{2\log^6n\log^2p_n} \geq \frac{(w - 2)^4}{4\log^8n}.
\end{displaymath}
\end{lemma}

Now we give a proof of Proposition \ref{prop401}.

\begin{proof}[Proof of Proposition \ref{prop401}]
Let $n \geq \max \{N_0, K_1, K_2, 3520\}$. By \cite[Theorem 2]{ca2017}, we have 
\begin{equation}
p_n &gt; n \left( \log p_n - 1 - \frac{1}{\log p_n} - \frac{3.15}{\log^2p_n} - \frac{12.85}{\log^3p_n} - \frac{71.3}{\log^4p_n} - \frac{463.2275}{\log^5p_n} - 
\frac{4585}{\log^6p_n} \right). \tag{5.6} \label{5.6}
\end{equation}
For convenience, we write $w = \log \log n$, $y = \log n$, and $z = \log p_n$. From Corollary \ref{kor206}, it follows that
\begin{equation}
- \frac{1}{z} \geq \Psi_1(n) = - \frac{1}{y} + \frac{w}{y^2} - \frac{w^2-w+1}{y^2z} - \frac{P_5(w)}{2y^3z} + \frac{P_6(w)}{2y^4z}. \tag{5.7} \label{5.7}
\end{equation}
Similarly to the proof of \eqref{3.10}, we use Proposition \ref{prop204} to get
\begin{equation}
- \frac{1}{z^2} \geq \Psi_2(n), \tag{5.8} \label{5.8}
\end{equation}
where
\begin{displaymath}
\Psi_2(n) = - \frac{1}{y^2} + \frac{w}{y^3} + \frac{w}{y^2z} - \left( \frac{1}{y} + \frac{1}{z} \right) \left(\frac{w^2 - w + 1}{y^2z} + \frac{P_5(w)}{2y^3z} - 
\frac{1}{2z} \sum_{k=4}^6\frac{P_{k+5}(w)}{y^k} \right).
\end{displaymath}
Using $P_8(\log \log x) \geq 0$ for every $x \geq 3$, $P_7(\log \log x) \geq 0$ for every $x \geq 3520$, and Corollary \ref{kor205}, we get
\begin{align}
- \frac{1}{z^3} \geq \Psi_3(n) &amp; = - \frac{1}{y^3} + \frac{w}{y^4} + \frac{w}{y^3z} + \frac{w}{y^2z^2} - \frac{w^2 - w + 1}{y^4z} - \frac{w^2 - w + 1}{y^3z^2} -
\frac{w^2 - w + 1}{y^2z^3} \nonumber \\
&amp; \p{\q\q} - \frac{P_5(w)}{2y^5z} - \frac{P_5(w)}{2y^4z^2} - \frac{P_5(w)}{2y^3z^3} + \frac{P_6(w)}{2y^6z} + \frac{P_6(w)}{2y^5z^2} + \frac{P_6(w)}{2y^4z^3} +
\frac{P_7(w)}{2y^7z}. \tag{5.9} \label{5.9}
\end{align}
By \eqref{5.1}, $3.15 - (B_6 + B_7 + B_8 + B_9 + B_{10}) \geq 0$. Since $n \geq N_0$ is assumed, we have $F_0(n) \geq 0$. Hence, by \eqref{3.12} and 
\eqref{5.5}, we see that
\begin{equation}
\frac{d(n)}{2y^2} \leq G_1(w) - \frac{a_1(n)}{2y^3} + \frac{(3.15 - (B_6 + B_7 + B_8 + B_9 + B_{10}))w}{y^2z}, \tag{5.10} \label{5.10}
\end{equation}
where $d(n) = 11.3 - b_1(n)$. We have $n \geq K_1$. This means that $\sum_{i=1}^{10} H_i(n) \geq 0$. So we can add $\sum_{i=1}^{10} H_i(n)$ to the right-hand 
side of \eqref{5.10} and use Lemma \ref{lem402} to get
\begin{align*}
\frac{d(n)}{2y^2} &amp; \leq G_1(w) - \frac{a_1(n)}{2y^3} + 12.85 \left( \Psi_3(n) + \frac{1}{y^3} - \frac{w}{y^4} + \frac{P_5(w)}{2y^5z} 
- \frac{P_6(w)}{2y^5z^2} \right) \\
&amp; \p{\q\q} + 3.15 \left( \Psi_2(n) + \frac{1}{y^2} - \frac{w}{y^3} + \frac{w^2 - w +1}{y^3z} - \frac{P_6(w)}{2y^5z} - \frac{P_7(w)}{2y^5z^2} \right) \\
&amp; \p{\q\q} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} - \frac{4585}{z^6} - \frac{(w^2-w+1)^2}{y^4z} + \frac{P_6(w)}{2y^4z} + \frac{P_6(w)P_9(w)}{4y^7z} \\
&amp; \p{\q\q} + \frac{P_8(w)}{2y^6z} - \frac{P_{10}(w)}{2y^5z} + \frac{P_{11}(w)}{2y^5z^2} + \frac{P_{12}(w)}{4y^6z} - \frac{(w - 2)^4}{4y^8},
\end{align*}
where $\Psi_2(n)$ and $\Psi_3(n)$ are given as in \eqref{5.8} and \eqref{5.9}, respectively. Applying the defining formulas of $P_{10}, P_{11}, P_{12}$, 
and $G_1$ to the last inequality, we find
\begin{align*}
\frac{d(n)}{2y^2} &amp; \leq - \frac{a_1(n)}{2y^3} + \frac{w^2-w+1}{y^2} \cdot \Psi_1(n) + \frac{P_9(w)}{2y^3} \cdot \Psi_1(n) + 12.85 \Psi_3(n) + 
\frac{(w-1)^2}{2y^2} \\
&amp; \p{\q\q} - \frac{w^2-6w}{2y^3} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} - \frac{4585}{z^6} + 3.15 \left( \Psi_2(n) + \frac{1}{y^2} + \frac{w^2 - w +1}{y^3z} 
\right) \\
&amp; \p{\q\q} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( \frac{w-1}{y} + \frac{w-2}{y^2} \right)^k + \sum_{k=4}^6 \frac{P_{k+5}(w)}{2y^kz},
\end{align*}
where $\Psi_1(n)$ is given as in \eqref{5.7}. Note that $w^2-w+1$ and $P_9(w)$ are nonnegative. Therefore, we can apply \eqref{5.7} and \eqref{5.9} to 
the last inequality and get
\begin{align*}
\frac{d(n)}{2y^2} &amp; \leq - \frac{a_1(n)}{2y^3} - \frac{w^2-w+1}{y^2z} - \frac{P_9(w)}{2y^3z} - \frac{12.85}{z^3} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} - 
\frac{4585}{z^6} \\
&amp; \p{\q\q} + \frac{(w-1)^2}{2y^2} - \frac{w^2-6w}{2y^3} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( \frac{w-1}{y} + \frac{w-2}{y^2} \right)^k \\
&amp; \p{\q\q} + 3.15 \left( \Psi_2(n) + \frac{1}{y^2} + \frac{w^2 - w +1}{y^3z} \right) + \sum_{k=4}^6 \frac{P_{k+5}(w)}{2y^kz}.
\end{align*}
Since $P_9(x) = P_5(x) + 2 \cdot 3.15(x^2-x+1)$ and $d(n) = 11.3 - b_1(n)$, the last inequality is equivalent to
\begin{align*}
\frac{5-b_1(n)}{2y^2} &amp; \leq - \frac{a_1(n)}{2y^3} - \frac{w^2-w+1}{y^2z} + \frac{(w-1)^2}{2y^2} - \frac{w^2-6w}{2y^3} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( 
\frac{w-1}{y} + \frac{w-2}{y^2} \right)^k \\
&amp; \p{\q\q} + 3.15 \Psi_2(n) - \frac{12.85}{z^3} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} - \frac{4585}{z^6} - \frac{P_5(w)}{2y^3z} + \sum_{k=4}^6 
\frac{P_{k+5}(w)}{2y^kz}.
\end{align*}
Using \eqref{5.8} and Proposition \ref{prop204}, we get the inequality
\begin{align*}
\frac{5-b_1(n)}{2y^2} &amp; \leq - \frac{1}{z} - \frac{3.15}{z^2} - \frac{12.85}{z^3} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} -  \frac{4585}{z^6} + \frac{1}{y} - 
\frac{w}{y^2} + \frac{(w-1)^2}{2y^2} \\
&amp; \p{\q\q} - \frac{w^2-6w}{2y^3} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( \frac{w-1}{y} + \frac{w-2}{y^2} \right)^k  - \frac{a_1(n)}{2y^3}
\end{align*}
which is equivalent to
\begin{align}
\frac{w-2}{y} &amp; \leq \frac{w-1}{y} + \frac{w-2}{y^2} - \frac{w^2-6w + a_1(n)}{2y^3} - \sum_{k=2}^4 \frac{(-1)^k}{k} \left( \frac{w-1}{y} + \frac{w-2}{y^2} 
\right)^k \nonumber \\
&amp; \p{\q\q} + \frac{w^2-6w+b_1(n)}{2y^2} - \frac{1}{z} - \frac{3.15}{z^2} - \frac{12.85}{z^3} - \frac{71.3}{z^4} - \frac{463.2275}{z^5} - \frac{4585}{z^6}. 
\tag{5.11} \label{5.11}
\end{align}
Since $\log(1+t) \geq \sum_{k=1}^4 (-1)^{k+1}t^k/k$ for every $t &gt; -1$ and both $g_1(x) = -x^2/2 + x^3/3$ and $g_2(x) = -x^4/4$ are decreasing on the interval 
$[0,1]$, we can use \eqref{5.2} and \eqref{5.3} to see that the inequality \eqref{5.11} implies
\begin{align*}
\frac{w-2}{y} - \frac{w^2 - 6w + b_1(n)}{2y^2} &amp; \leq \log \left(1 + \frac{w-1}{y} + \frac{w - 2}{y^2} - \frac{w^2-6w + a_1(n)}{2y^3} \right) - \frac{1}{z} - 
\frac{3.15}{z^2} \\
&amp; \p{\q\q} - \frac{12.85}{z^3} - \frac{71.3}{z^4}- \frac{463.2275}{z^5} - \frac{4585}{z^6}.
\end{align*}
Now we add $y+w-1$ to both sides of the last inequality und use \eqref{5.5} to get
\begin{displaymath}
y + w - 1 + \frac{w - 2}{y} - \frac{w^2 - 6w + b_1(n)}{y} \leq z -1 - \frac{1}{z} - \frac{3.15}{z^2} - \frac{12.85}{z^3} - \frac{71.3}{z^4} - 
\frac{463.2275}{z^5} - \frac{4585}{z^6}.
\end{displaymath}
Finally, we multiply the last inequality by $n$ and apply \eqref{5.6} to complete the proof.
\end{proof}

Now, we give a proof of Theorem \ref{thm104}.

\begin{proof}[Proof of Theorem \ref{thm104}]
Clearly, Theorem \ref{thm103} implies the validity of the inequality \eqref{1.14} for every integer $n$ satisfying $2 \leq n \leq \pi(10^{19})$. Next, we prove 
the inequality \eqref{1.14} for every $n \geq M_0$, where $M_0 = \pi(10^{19}) + 1 = 234\,057\,667\,276\,344\,608$. In order to do this, let $A_0 = 0.914$. Then, 
similar to the proof of Lemma \ref{lem303}, we get $\log n \geq 0.914 \log p_n$ for every integer $n \geq M_0$. So can chose $N_0 = M_0$. In the following table 
we give explicit values for $B_i$:
\vspace{2mm}
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
  $i$ &amp;     1&amp;     2&amp;    3&amp;     4&amp;     5&amp;     6&amp;      7&amp;      8&amp;     9&amp;     10 \\ \hline
$B_i$ &amp; 0.132&amp; 3.021&amp; 1.11&amp; 0.023&amp; 1.993&amp; 0.055&amp; 0.0006&amp; 0.0199&amp; 0.055&amp; 0.0125 \\ \hline
\end{tabular}
\end{center}
\vspace{2mm}
Then $H_i(n) \geq 0$ for every $n \geq M_0$ and each integer $i$ satisfying $1 \leq i \leq 10$. So we can set $K_1 = M_0$. The proof that $H_i(n) \geq 0$ for 
every $n \geq M_0$ and each integer $i$ with $1 \leq i \leq 10$ can be found in Section 7. Furthermore, the above table indicates
\begin{equation}
3.15 - (B_6 + B_7 + B_8 + B_9 + B_{10}) = 3.007. \tag{5.12} \label{5.12}
\end{equation}

\textit{Step 1}. We set $a_1(n) = 0.2 y - w^2 + 6w$. Then, by \eqref{1.11} and \eqref{5.2}--\eqref{5.5}, we can choose $K_2 = 33$. Using \eqref{5.5} and 
\eqref{5.12}, we obtain
\begin{align*}
b_1(n) &amp; = 11.5 - \frac{2w^3-18w^2+65.390388w-97.1}{3y} + \rho(n),
\end{align*}
where
\begin{align}
\rho(n) &amp; = \frac{w^4-12w^3+46.6w^2-112w+40}{2y^2} + \frac{2w^4-21.3w^3+40.3w^2-41.5w+12}{y^3} \nonumber \\
&amp; \p{\q\q} + \frac{9w^4-56w^3+129w^2-132w+52}{3y^4} + \frac{2w^4-14w^3+36w^2-40w+16}{y^5}. \tag{5.13} \label{5.13}
\end{align}
In this step, we show that $b_1(n) \leq 11.5$ for every $n \geq M_0$. For this purpose, we set
\begin{align*}
\alpha(x,t) &amp; = 2(2x^3 + 65.390388x)e^{4x} - 2(18t^2+97.1)e^{4t} \\
&amp; \p{\q\q} + 3(12x^3 + 112x)e^{3x} - 3(t^4 + 46.6t^2 + 40)e^{3t} \\
&amp; \p{\q\q} + 6(21.3x^3 + 41.5x)e^{2x} - 6(2t^4 + 40.3t^2 + 12)e^{2t} \\
&amp; \p{\q\q} + 2(56x^3 + 132x)e^x  - 2(9t^4 + 129t^2 + 52)e^t \\
&amp; \p{\q\q} + 6(14x^3 + 40x) - 6(2t^4 + 36t^2 + 16).
\end{align*}
Note that this function satisfies the identity 
\begin{equation}
\alpha(w,w) = 6(11.5 - b_1(n))y^5. \tag{5.14} \label{5.14}
\end{equation}
If $t_0 \leq x \leq t_1$, then $\alpha(x,x) \geq \alpha(t_0,t_1)$. We check with a computer that $\alpha(3.6 + i \cdot 10^{-3}, 3.6 + (i+1) \cdot 10^{-3}) 
\geq 0$ for every integer $i$ satisfying $0 \leq i \leq 5399$. Hence by \eqref{5.14}, 
\begin{equation}
b_1(n) \leq 11.5 \q\q (3.6 \leq w \leq 9). \tag{5.15} \label{5.15}
\end{equation}
Next, we show that $\alpha(x,x) \geq 0$ for every $x \geq 9$. Since $2(2x^3-18x^2+65.390388x-97.1) \geq 982$ for every $x \geq 9$, we have
\begin{align*}
\alpha(x,x) &amp; \geq 982e^{4x} - 3(x^4-12x^3+46.6x^2-112x+40)e^{3x} \\
&amp; \p{\q\q} - 6(2x^4-21.3x^3+40.3x^2-41.5x+12)e^{2x} \\
&amp; \p{\q\q} - 2(9x^4-56x^3+129x^2-132x+52)e^x \\
&amp; \p{\q\q} - 6(2x^4-14x^3+36x^2-40x+16).
\end{align*}
Note that $982e^{x} - 3(x^4-12x^3+46.6x^2-112x+40) \geq 7\,955\,369$ for every $x \geq 9$. Therefore, $\alpha(x,x) \geq 0$ for every $x \geq 9$. Combined with 
\eqref{5.14} and \eqref{5.15}, it gives $b_1(n) \leq 11.5$ for every $n \geq M_0 &gt; \exp(\exp(3.6))$. Applying this to Proposition \ref{prop401}, we get
\begin{displaymath}
p_n &gt; n \left( y + w - 1 + \frac{w - 2}{y} - \frac{w^2-6w + 11.5}{2y^2} \right)
\end{displaymath}
for every $n \geq M_0$.

\textit{Step 2}. We set $a_1(n) = 11.5$. Then $K_2 = 47$ is a suitable choice for $K_2$. Combined with \eqref{5.5} and \eqref{5.12}, it gives
\begin{displaymath}
b_1(n) = 11.3 - \frac{2w^3 - 21w^2 + 83.390388w - 131.6}{3y} + \rho(n),
\end{displaymath}
where $\rho(n)$ is defined as in \eqref{5.13}. We set
\begin{align*}
\beta(x,t) &amp; = 0.15e^{5x} + 2(2x^3 + 83.390388x)e^{4x} - 2(21t^2 + 131.6)e^{4t} \\
&amp; \p{\q\q} + 3(12x^3 + 112x)e^{3x} - 3(t^4 + 46.6t^2 + 40)e^{3t} \\
&amp; \p{\q\q} + 6(21.3x^3 + 41.5x)e^{2x} - 6(2t^4 + 40.3t^2 + 12)e^{2t} \\
&amp; \p{\q\q} + 2(56x^3 + 132x)e^x  - 2(9t^4 + 129t^2 + 52)e^t \\
&amp; \p{\q\q} + 6(14x^3 + 40x) - 6(2t^4 + 36t^2 + 16).
\end{align*}
Then $\beta(w,w) = 6(11.325 - b_1(n))y^5$. Similarly to the first step, we get
\begin{displaymath}
b_1(n) \leq 11.325 \q\q (3.686 \leq w \leq 7).
\end{displaymath}
Therefore, it suffices to verify that $\beta(x,x) \geq 0$ for every $x \geq 7$. Notice that $0.15e^x + 2(2x^3-21x^2+83.390388x-131.6) \geq 382$ for every $x 
\geq 7$. Thus we get
\begin{align*}
\beta(x,x) &amp; \geq 382e^{4x} - 3(x^4-12x^3+46.6x^2-112x+40)e^{3x} \\
&amp; \p{\q\q} - 6(2x^4-21.3x^3+40.3x^2-41.5x+12)e^{2x} \\
&amp; \p{\q\q} - 2(9x^4-56x^3+129x^2-132x+52)e^x \\
&amp; \p{\q\q} - 6(2x^4-14x^3+36x^2-40x+16).
\end{align*}
Since $382e^x - 3(x^4-12x^3+46.6x^2-112x+40) \geq 419\,440$ for every $x \geq 7$, we conclude that $\beta(x,x) \geq 0$ for every $x \geq 7$. Hence $b_1(n) 
\leq 11.325$ for every $n \geq M_0 &gt; \exp(\exp(3.686))$. So, by Proposition \ref{prop401},
\begin{displaymath}
p_n &gt; n \left( y + w - 1 + \frac{w - 2}{y} - \frac{w^2-6w + 11.325}{2y^2} \right)
\end{displaymath}
for every $n \geq M_0$.

\textit{Step 3}. Here we set $a_1(n) = 11.325$. Then we can choose $K_2 = 47$. By \eqref{5.5} and \eqref{5.12},
\begin{displaymath}
b_1(n) = 11.3 - \frac{2w^3 - 21w^2 + 83.390388w - 131.075}{3y} + \rho(n),
\end{displaymath}
where $\rho(n)$ is defined as in \eqref{5.13}. To show that $b_1(n) \leq 11.321$ for every $n \geq M_0$, we set
\begin{align*}
\gamma(x,t) &amp; = 0.126e^{5x} + 2(2x^3 + 83.390388x)e^{4x} - 2(21t^2 + 131.075)e^{4t} \\
&amp; \p{\q\q} + 3(12x^3 + 112x)e^{3x} - 3(t^4 + 46.6t^2 + 40)e^{3t} + 6(21.3x^3 + 41.5x)e^{2x} \\
&amp; \p{\q\q} - 6(2t^4 + 40.3t^2 + 12)e^{2t} + 2(56x^3 + 132x)e^x  - 2(9t^4 + 129t^2 + 52)e^t \\
&amp; \p{\q\q} + 6(14x^3 + 40x) - 6(2t^4 + 36t^2 + 16).
\end{align*}
Notice that $\gamma(w,w) = 6(11.321 - b_1(n))y^5$. Analogously to the first step, we obtain $b_1(n) \leq 11.321$ for $w$ satisfying $3.68 \leq w \leq 7$. Next 
we find $b_1(n) \leq 11.321$ for $w \geq 7$. Note that $0.126e^x + 2(2x^3-21x^2+83.390388x-131.075) \geq 357.491$ for every $x \geq 7$. Therefore,
\begin{align*}
\gamma(x,x) &amp; \geq 357e^{4x} - 3(x^4-12x^3+46.6x^2-112x+40)e^{3x} \\
&amp; \p{\q\q} - 6(2x^4-21.3x^3+40.3x^2-41.5x+12)e^{2x} \\
&amp; \p{\q\q} - 2(9x^4-56x^3+129x^2-132x+52)e^x \\
&amp; \p{\q\q}- 6(2x^4-14x^3+36x^2-40x+16).
\end{align*}
Since $357e^{x} - 3(x^4-12x^3+46.6x^2-112x+40) \geq 392\,024$ for every $x \geq 7$, we get $\gamma(x,x) \geq 0$ for every $x \geq 7$. So $b_1(n) \leq 
11.321$ for every $n \geq M_0 &gt; \exp(\exp(3.68))$. Now Proposition \ref{prop401} implies the required inequality for every $n \geq M_0$ which completes the 
proof.
\end{proof}

Denoting the right-hand side of \eqref{1.11} by $D_{\rm low}(n)$ and the right-hand side of \eqref{1.14} by $A_{low}(n)$, we use \seqnum{A006988} to compare the 
error term of the approximation from Theorem \ref{thm104} with the approximation from \eqref{1.11} for the $10^n$th prime number:
\begin{center}
\begin{tabular}{|c||r|r|r|}
\hline
$n$\rule{0mm}{4mm}         &amp; $p_n$ &amp; $\lceil p_n - D_{\rm low}(n)\rceil  $ &amp; $\lceil p_n - A_{low}(n) \rceil$ \\ \hline
$ 10^{10} $\rule{0mm}{4mm} &amp; $252\,097\,800\,623$ &amp; $22\,918\,665$ &amp; $1\,553\,620$ \\\hline
$ 10^{11} $\rule{0mm}{4mm} &amp; $2\,760\,727\,302\,517$ &amp; $221\,928\,766$ &amp; $12\,203\,725$ \\\hline
$ 10^{12} $\rule{0mm}{4mm} &amp; $29\,996\,224\,275\,833$ &amp; $2\,149\,187\,973$ &amp; $116\,712\,205$ \\\hline
$ 10^{13} $\rule{0mm}{4mm} &amp; $323\,780\,508\,946\,331$ &amp; $20\,674\,500\,003$ &amp; $1\,107\,237\,510$ \\\hline
$ 10^{14} $\rule{0mm}{4mm} &amp; $3\,475\,385\,758\,524\,527$ &amp; $198\,184\,329\,536$ &amp; $10\,418\,290\,134$ \\\hline
$ 10^{15} $\rule{0mm}{4mm} &amp; $37\,124\,508\,045\,065\,437$ &amp; $1\,896\,434\,754\,032$ &amp; $97\,120\,372\,631$ \\\hline
$ 10^{16} $\rule{0mm}{4mm} &amp; $394\,906\,913\,903\,735\,329$ &amp; $18\,139\,062\,711\,550$ &amp; $901\,415\,873\,097$ \\\hline
$ 10^{17} $\rule{0mm}{4mm} &amp; $4\,185\,296\,581\,467\,695\,669$ &amp; $173\,543\,282\,219\,005$ &amp; $8\,342\,526\,771\,836$ \\\hline
$ 10^{18} $\rule{0mm}{4mm} &amp; $44\,211\,790\,234\,832\,169\,331$ &amp; $1\,661\,592\,139\,340\,947$ &amp; $77\,153\,499\,580\,018$ \\\hline
$ 10^{19} $\rule{0mm}{4mm} &amp; $465\,675\,465\,116\,607\,065\,549$ &amp; $15\,924\,846\,933\,652\,812$ &amp; $713\,638\,559\,773\,813$ \\\hline
$ 10^{20} $\rule{0mm}{4mm} &amp; $4\,892\,055\,594\,575\,155\,744\,537$ &amp; $152\,800\,345\,036\,619\,338$ &amp; $6\,606\,690\,561\,425\,196$ \\ \hline
\end{tabular}
\vspace{2mm}
\end{center}

\begin{remark}
Compared to Theorem \ref{thm104}, the asymptotic expansion \eqref{1.2} implies a better lower bound for the $n$th prime number, which corresponds to 
the first five terms, namely that
\begin{equation}
p_n &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log\log n)^2-6\log \log n + 11}{2\log^2 n} \right) \tag{5.16} \label{5.16}
\end{equation}
for all sufficiently large values of $n$. Let $r_3$ denote the smallest positive integer such that the inequality \eqref{5.16} holds for every $n \geq r_3$. 
Under the assumption that the Riemann hypothesis is true, Arias de Reyna and Toulisse \cite[Theorem 6.4]{adrt} proved that $3.9 \cdot 10^{30} &lt; r_3 \leq 3.958 
\cdot 10^{30}$.
\end{remark}

\section{New estimates for $\vartheta(p_n)$}

Chebyshev's $\vartheta$-function is defined by
\begin{displaymath}
\vartheta(x) = \sum_{p \leq x} \log p,
\end{displaymath}
where $p$ runs over primes not exceeding $x$. Notice that the prime number theorem is equivalent to
\begin{equation}
\vartheta(x) \sim x \quad\quad (x \to \infty). \tag{6.1} \label{6.1}
\end{equation}
By proving the existence of a zero-free region for the Riemann zeta-function $\zeta(s)$ to the left of the line $\text{Re}(s) = 1$, de la Vall\'{e}e-Poussin
\cite{vallee1899} found an estimate for the error term in \eqref{6.1} by proving $\vartheta(x) = x + O(x e^{-c\sqrt{\log x}})$, where $c$ is a positive 
absolute constant. Applying \eqref{1.2} to the last asymptotic formula, we see that
\begin{displaymath}
\vartheta(p_n) = n \left( \log n + \log_2n - 1 + \frac{\log_2n - 2}{\log n} - \frac{(\log_2n)^2 - 6 \log_2n + 11}{2 \log^2 n} + O\left( 
\frac{(\log_2n)^3}{\log^3n} \right)\right),
\end{displaymath}
where $\log_2 n = \log \log n$. In this direction, many estimates for $\vartheta(p_n)$ were obtained (see for example Massias and Robin \cite[Th\'{e}or\`{e}me 
B]{mr}). The current best ones are due to Dusart \cite[Propositions 5.11 and 5.12]{dusart2017}. He found that
\begin{displaymath}
\vartheta(p_n) \geq n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2.04}{\log n} \right)
\end{displaymath}
for every $n \geq \pi(10^{15}) + 1 = 29\,844\,570\,422\,670$, and that the inequality
\begin{displaymath}
\vartheta(p_n) \leq n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{0.782}{\log^2 n} \right)
\end{displaymath}
holds for every $n \geq 781$. Using Theorems \ref{thm101} and \ref{thm104}, we find the following estimates for $\vartheta(p_n)$, which improve the estimates 
given by Dusart.

\begin{proposition} \label{prop601}
For every integer $n \geq 2$, we have
\begin{displaymath}
\vartheta(p_n) &gt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2 - 6 \log \log n + 11.621}{2 \log^2 n} \right), 
\end{displaymath}
and for every integer $n \geq 2581$, we have
\begin{displaymath}
\vartheta(p_n) &lt; n \left( \log n + \log \log n - 1 + \frac{\log \log n - 2}{\log n} - \frac{(\log \log n)^2 - 6 \log \log n + 10.367}{2 \log^2 n} \right).
\end{displaymath}
\end{proposition}

\begin{proof}
From \cite[Theorem 1]{ca2017}, it follows that
\begin{equation}
p_n - \frac{0.15p_n}{\log^3 p_n} &lt; \vartheta(p_n) &lt; p_n + \frac{0.15p_n}{\log^3 p_n}, \tag{6.2} \label{6.2}
\end{equation}
where the left-hand side inequality is valid for every integer $n \geq 841\,508\,302$ and the right-hand side inequality holds for every positive integer $n$. 
By Rosser and Schoenfeld \cite[Corollary 1]{rosserschoenfeld1962}, we have $n &gt; p_n/\log p_n$ for every $n \geq 7$. Applying the last inequality to the 
left-hand side inequality of \eqref{6.2}, we get $\vartheta(p_n) &gt; p_n - 0.15n/\log^2n$ for every $n \geq 841\,508\,302$. Now we apply Theorem \ref{thm104} to 
get the desired lower bound for $\vartheta(p_n)$ for every $n \geq 841\,508\,302$. By BÃ¼the \cite[Theorem 2]{buethe}, we have
\begin{equation}
\vartheta(x) \geq x - \frac{\sqrt{x}}{8\pi} \; \log^2 x \q\q (599 &lt; x \leq 1.89 \times 10^{21}). \tag{6.3} \label{6.3}
\end{equation}
Now we apply Theorem \ref{thm103} to \eqref{6.3} and get the required lower bound for $\vartheta(p_n)$ for every integer $n$ with $200\,000 \leq n \leq 
841\,508\,301$. We check the remaining cases for $n$ with a computer.

Similarly to the first part of the proof, we apply the inequality $n &gt; p_n/\log p_n$ to the right-hand side inequality of \eqref{6.2} to get $\vartheta(p_n) &lt; 
p_n + 0.15n/\log^2n$ for every $n \geq 7$. Now we use Theorem \ref{thm101} to get the required upper bound for $\vartheta(p_n)$ for every $n \geq 46\,254\,381$. 
For smaller values of $n$, we use a computer.
\end{proof}

\section{Appendix}

Let $M_0 = \pi(10^{19})+1 = 234\,057\,667\,276\,344\,608$. In the proof of Theorem \ref{thm104}, we note a table in which we give explicit values of $B_i$. In 
this appendix, we show that the $H_i$ defined at the start of paragraph 5 are non-negative for every integer $n \geq M_0$ for the given values of $B_i$. We 
start with the claim concerning $H_1$.

\begin{proposition} \label{prop701}
If $B_1 = 0.132$, then $H_1(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. We have $P_{11}(x) \geq 0$ for every $x \geq 0.6$ and $P_6(x) \geq 0$ for every $x \geq 0.6$. Using Lemma \ref{lem303}, we get
\begin{equation}
H_1(n) \geq \frac{f_1(\log \log n)}{4\log^6n\log p_n} \tag{7.1} \label{7.1}
\end{equation}
for every integer $n \geq M_0$, where $f_1(x) = 0.528 xe^{3x} - 2P_{10}(x)e^x + 1.74 P_{11}(x) + P_{12}(x) + 19.45233P_6(x)$. We 
show that $f(x) \geq 0$ for every $x \geq x_0$. For this, we set $g(x) = (57.024+42.768x)e^{2x} + (-24.6x^4 - 322.1x^3 - 1137.1x^2 - 1265.98x - 512.24)$. It 
is easy to show that $g(x) \geq 3 \cdot 10^5$ for every $x \geq x_0$. So, $f_1^{(4)}(x) = g(x)e^x + 240x - 1005.6 \geq 0$ for every $x \geq x_0$. Now it is 
easy to see that $f(x) \geq 0$ for every $x \geq x_0$. Applying this to \eqref{7.1}, we get $H_1(n) \geq 0$ for every integer $n \geq M_0$.
\end{proof}

Let $B_2 = 3.021$. Before we check that $H_2(n) \geq 0$ for every integer $n \geq M_0$, we introduce the following

\begin{definition}
For $x \geq 1$, let
\begin{displaymath}
\Phi(x) = e^x + x + \log \left( 1 + \frac{x-1}{e^x} + \frac{x-2.1}{e^{2x}} \right).
\end{displaymath}
\end{definition}

We note the following three properties of the function $\Phi(x)$.

\begin{lemma} \label{lem702}
For every $x \geq 0.179$, we have $\Phi'(x) \geq e^x + 3/4$.
\end{lemma}

\begin{proof}
We have $\Phi'(x) \geq e^x + 3/4$ if and only if $g(x) = e^{2x} - 3xe^x + 7e^x - 7x + 18.7 \geq 0$. Since $g''(x) = 4e^{2x} - (3x-1)e^x \geq 0$ for every $x 
\geq 0$ and $g'(0.179) \geq 0$, we obtain $g'(x) \geq 0$ for every $x \geq 0.179$. If we combine this with $g(0.179) \geq 26.6$, we get $g(x) \geq 0$ for 
every $x \geq 0.179$. 
\end{proof}

\begin{lemma} \label{lem703}
For every $x \geq 1.246$, we have $\Phi(x) \geq e^x + x$.
\end{lemma}

\begin{proof}
The desired inequality holds if and only if $(x-1)e^x +  x - 2.1 \geq 0$. Since the last inequality holds for every $x \geq 1.246$, we arrived at the end of 
the proof.
\end{proof}

\begin{lemma} \label{lem704}
For every integer $n \geq 3$, we have $\Phi( \log \log n) \leq \log p_n$.
\end{lemma}

\begin{proof}
The claim follows directly from \eqref{1.11}.
\end{proof}

Next, we use these properties to see that $H_2(n) \geq 0$ for every integer $n \geq M_0$.

\begin{proposition} \label{prop705}
Let $B_2 = 3.021$. Then $H_2(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. We set $f_2(x) = 3.021x \Phi^3(x) + 12.85xe^x \Phi^2(x) - 71.3e^{3x}$ and use Lemmata \ref{lem702} and \ref{lem703} to obtain
\begin{align*}
f_2'(x) &amp; \geq 3.021(e^x + x)^3 + 21.913xe^x(e^x+x)^2 + 12.85e^x(e^x+x)^2 \nonumber \\
&amp; \p{\q\q} + 25.7xe^{2x}(e^x+x)-213.9e^{3x} \tag{7.2} \label{7.2}
\end{align*}
for every $x \geq 1.25$. We denote the right-hand side of the last inequality by $g_2(x)$. A straightforward calculation gives $g_2^{(3)}(x) \geq (1285.551x - 
4061.232)e^{3x} \geq 0$ for every $x \geq x_0$. Now it is easy to see that $g_2(x) \geq 0$ for every $x \geq x_0$. Applying this to \eqref{7.2}, we see that 
$f_2'(x) \geq 0$ for every $x \geq x_0$. Since $f_2(x_0) \geq 268.5$, we obtain $f_2(\log \log n) \geq 0$ for every integer $n \geq M_0$. Finally, we apply 
Lemma \ref{lem704}.
\end{proof}

\begin{proposition} \label{prop706}
If $B_3 = 1.11$, then $H_3(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$ and let $f_3(x) = 2.22 x \Phi(x) - 35.15x^2 + 44.6x - 42.08$. Using Lemmata \ref{lem702} and \ref{lem703}, we get $f_3'(x) \geq 
(2.22e^x - 65.86)x \geq 0$ holds for every $x \geq x_0$. Combined with $f_3(x_0) \geq 2.42$ and Lemma \ref{lem704}, we get that $H_3(n) \geq 0$ for every 
integer $n \geq M_0$.
\end{proof}

\begin{proposition} \label{prop707}
Let $B_4 = 0.023$. Then $H_4(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. We set $f_4(x) = 0.046xe^x \Phi(x) + 3.15x^3 - 57.45x^2 + 113.01x - 80.05$ and have $f_4(M_0) \geq 10.103$. By Lemmata \ref{lem702} 
and \ref{lem703}, we get $f_4'(x) \geq (0.046(e^x(e^x+x) + e^{2x}) + 9.45x - 114.9)x \geq 0$ for every $x \geq x_0$. Hence $f_4(\log \log n) \geq 0$ for every 
integer $n \geq M_0$ and we can apply Lemma \ref{lem704}.
\end{proof}

\begin{proposition} \label{prop708}
Let $B_5 = 1.993$. Then we have $H_5(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. To proof the claim, we define $f_5(x) = 3.986xe^x - 2x^4 + 5x^3 - 47.15x^2 + 60x - 48.28$. Since $f_5'''(x) \geq 0$ for every $x 
\geq x_0$ and $f_5''(x_0) \geq 0$, we obtain $f_5''(x) \geq 0$ for every $x \geq x_0$. Combined with $f_5'(x_0) \geq 0$, it turns out that $f_5'(x) \geq 0$ for 
every $x \geq x_0$. Together with $f_5(x_0) \geq 0.203$, we conclude that $f_5(\log \log n) \geq 0$, and thus $H_5(n) \geq 0$, for every integer $n \geq M_0$. 
\end{proof}

Adding the constants $B_1, \ldots, B_5$ given in Proposition \ref{prop701} and Propositions \ref{prop705}-\ref{prop708}, we get $12.85 - B_1 - B_2 - B_3 - 
B_4 - B_5 = 6.571$. Now we set $B_6 = 0.055$ to obtain the following result.

\begin{proposition} \label{prop709}
Let $B_6 = 0.055$. Then $H_6(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. Furthermore, let $r(x,t) = (0.118e^x + 4.116)x \Phi(x) + 3.15xe^x - 3.15(t^2+1)e^t$ and let $f_6(x) = r(x,x)$. If $t_0 \leq x \leq 
t_1$, then $f_6(x) \geq r(t_0,t_1)$. We 
check with a computer that $r(3.6 + i \cdot 10^{-3}, 3.6 + (i+1) \cdot 10^{-3}) \geq 0$ for every integer $i$ such that $0 \leq i \leq 599$. Hence $f_6(x) 
\geq 0$ for every $x$ such that $3.6 \leq x \leq 4.2$. To show that $f_6(x) \geq 0$ for every $x \geq 4.2$, we set
\begin{displaymath}
g(x) = (0.055(xe^x + e^x) + 6.571)(e^x+x) + (0.055e^x + 6.571)xe^x - 3.15xe^x(1+x).
\end{displaymath}
Then $g'(x) = h(x)e^x + 6.571$ where $h(x) = 0.22(1+x)e^x - 3.095x^2 - 2.714x + 10.047$. Since $h(x) \geq 0$ for every $x \geq 4.2$, we get $g'(x) \geq 0$ for 
every $x \geq 4.2$. Together with $g(4.2) \geq 0$, we see that $g(x) \geq 0$ for every $x \geq 4.2$. Using Lemmata \ref{lem702} and \ref{lem703}, we obtain 
$f_6'(x) \geq g(x) \geq 0$ for every $x \geq 4.2$. Combined with $f_6(4.2) \geq 17.047$, we have $f_6(x) \geq 0$ for every $x \geq 4.2$.  Hence $f_6(x) 
\geq 0$ for every $x \geq x_0 \geq 3.6$. Now we apply Lemma \ref{lem704} to get $H_6(n) \geq 0$ for every integer $n \geq M_0$.
\end{proof}

\begin{proposition} \label{prop710}
If $B_7 = 0.0006$, then we have $H_7(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. Substituting the definition of $P_5(x)$, we get
\begin{displaymath}
H_7(n) = \frac{0.0006w}{y^2z} - \frac{38.55w^2 - 77.1w + 66.82}{2y^3z^3}.
\end{displaymath}
To show that $H_7(n) \geq 0$ for every integer $n \geq M_0$, we first consider the function $f_7(x) = 0.0012xe^x \Phi^2(x) - 38.55x^2 + 77.1x - 66.82$. We have 
$f_7(x_0) \geq 31.88$. Additionally, we use Lemmata \ref{lem702} and \ref{lem703} to get $f_7'(x) \geq (0.0012(e^x+x)^2(1+e^x) + 0.0024e^{2x}(e^x + x) - 77.1)x 
\geq 0$ for every $x \geq x_0$. Hence, $f_7(\log \log n) \geq 0$ for every integer $n \geq M_0$. Finally, it suffices to apply Lemma \ref{lem704}.
\end{proof}

\begin{proposition} \label{prop711}
Let $B_8 = 0.0199$. Then $H_8(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. We set $f_8(x) = 0.0199x\, \Phi^2(x) - 12.85(x^2-x+1)$. We have $f_8(x_0) \geq 0.906$. By Lemmata \ref{lem702} and \ref{lem703}, we 
obtain $f_8'(x) \geq (0.0199(e^x + x) + 0.0398 (e^x + x)e^x - 25.7)x \geq 0$ for every $x \geq x_0$. Hence $f_8(\log \log n) \geq 0$ for every integer $n \geq 
M_0$. Finally, we use Lemma \ref{lem704}.
\end{proof}

\begin{proposition} \label{prop712}
If $B_9 = 0.055$, then $H_9(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$. We define $f_9(x) = 0.055x\,\Phi^4(x) - 463.2275e^{2x}$. By Lemmata \ref{lem702} and \ref{lem703}, we have $f_9'(x) \geq ((0.055 + 
0.22x)(e^x+x)^2 - 926.455)e^{2x} \geq 0$ for every $x \geq x_0$. Combined with $f_9(x_0) \geq 2263.343$, we get $f_9(x) \geq 0$ for every $x \geq x_0$. 
Substituting $x = \log \log n$ in $f_9(x)$, we apply Lemma \ref{lem704} to see that $H_9(n) \geq 0$ for every integer $n \geq M_0$.
\end{proof}

Finally, we set $B_{10} = 0.0125$ and check that $H_{10}(n) \geq 0$ for every integer $n \geq M_0$.

\begin{proposition} \label{prop713}
Let $B_{10} = 0.0125$. Then we have $H_{10}(n) \geq 0$ for every integer $n \geq M_0$.
\end{proposition}

\begin{proof}
Let $x_0 = \log \log M_0$ and let $f_{10}(x) = 0.0125x \,\Phi^5(x) - 4585e^{2x}$. Applying Lemmata \ref{lem702} and \ref{lem703}, we get $f_{10}'(x) \geq 
(0.4x(e^{x}+x)^3 - 9170)e^{2x} \geq 0$ for every $x \geq x_0$. Together with $f_{10}(x_0) \geq 55867.822$, we see that $f_{10}(\log \log n) \geq 0$ for 
every integer $n \geq M_0$. Now, we use Lemma \ref{lem704} to conclude that $H_{10}(n) \geq 0$ for every integer $n \geq M_0$. 
\end{proof}

\section{Acknowledgment}

I would like to thank the anonymous referees for useful comments to improve the quality of this paper. Furthermore, I would like to thank R. for being a 
never-ending inspiration. 

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N05; Secondary 11A41.

\noindent \emph{Keywords:}
Chebyshev's $\vartheta$-function, prime counting function, prime number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A000040} and \seqnum{A006988}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 12 2017; revised versions received
March 15 2018; February 18 2019; February 25 2019; May 21 2019.
Published in {\it Journal of Integer Sequences}, May 23 2019.

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\noindent
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