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\begin{center}
\vskip 1cm{\LARGE\bf
Functions of Slow Increase\\
\vskip .1in
and Integer Sequences}
\vskip 1cm
\large
Rafael Jakimczuk\\
Divisi\'on Matem\'atica \\
Universidad Nacional de Luj\'an\\
Buenos Aires\\
Argentina\\
\href{mailto:jakimczu@mail.unlu.edu.ar}{\tt jakimczu@mail.unlu.edu.ar}\\
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\vskip .2in

\begin{abstract}
We study some properties of functions that satisfy the condition
$f'(x)=o\left(\frac{f(x)}{x}\right)$,
for $ x\rightarrow \infty $,  i.e.,
$\lim_{x\rightarrow \infty}\frac{ f'(x)}{\frac{f(x)}{x}}= 0$.
We call these ``functions of slow increase'',
since they satisfy the condition
$\lim_{x\rightarrow \infty}\frac{f(x)}{x^{\alpha}} =0$
for all $\alpha&gt;0$.
A typical example of a function of slow increase is the function
$f(x)= \log x$.
As an  application, we obtain some general results on sequence $A_n$ of
positive integers that satisfy the asymptotic formula $A_n
\sim n^s f(n)$, where $f(x)$ is a function of slow increase.
\end{abstract}

\newtheorem{theorem}{Theorem}
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\newtheorem{defin}[theorem]{Definition}
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\section{Functions of Slow Increase}

\begin{definition}\label{D1} 
Let $f(x)$ be a function defined on the interval  $[a, \infty)$ such that $f(x)&gt;0$, 
$\lim_{x\rightarrow \infty} f(x)=\infty $ and with continuous derivative $f'(x)&gt;0$.
The function $f(x)$ is {\em of slow increase} if the following condition holds.
\begin{equation}\label{A1} 
\lim_{x\rightarrow \infty}\frac{ f'(x)}{\frac{f(x)}{x}}= 0.
\end{equation}
\end{definition} 

Typical functions of slow increase are  $f(x)=\log x$, $f(x)=\log^2 x$, $f(x)=\log\log x$, $f(x)=\frac{\log x}{\log\log x}$ and $\Psi:(0,\infty)\rightarrow (0,\infty)$, $\Psi(x)= \frac{\Gamma'(x)}{\Gamma(x)}$, where $\Gamma(x)=\int^{\infty}_{0}t^{x-1}e^{-t}\ dt$, which generalize the harmonic sum $H_n: N^{*} \rightarrow R$, $H_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ to $(0,\infty)$, namely $H_n=\Psi(n+1)+\gamma$, where $\gamma$ is Euler's constant. 

We have the following theorems.

\begin{theorem}\label{T2} 
If $f(x)$ and $g(x)$ are functions of slow increase and  $C$ and $\alpha$ are positive constants  then the following functions  are  of slow increase.
\[f(x)+C, \qquad f(x)-C, \qquad C f(x),\qquad f(x)g(x),\qquad  f(x)^{\alpha},\]
\[f(g(x)),\qquad \log f(x),\qquad f(x^{\alpha}),\qquad f(x^{\alpha}g(x)),\qquad f(x)+g(x).\]
If $f(x)$ and $g(x)$ are functions of slow increase, $\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=\infty$ and $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)&gt;0$ then 
$\frac{f(x)}{g(x)}$ is  a function of slow increase.

If $h(x)$ is a function such that $h(x)&gt;0$, 
$\lim_{x\rightarrow \infty} h(x)=\infty $ and with continuous derivative $h'(x)&gt;0$, then 
$h(\log x)$ is a function of slow increase if and only if $\lim_{x\rightarrow \infty}\frac{h'(x)}{h(x)}=0$.

If $h(x)$ is a function such that $h(x)&gt;0$, 
$\lim_{x\rightarrow \infty} h(x)=\infty $ and with continuous derivative $h'(x)&gt;0$, then 
$e^{h(x)}$ is a function of slow increase if and only if $\lim_{x\rightarrow \infty}x h'(x)=0$.

If $f(x)$ is a function of slow increase the following limit holds.
\begin{equation}\label{A2}
\lim_{x\rightarrow \infty}\frac{\log f(x)}{\log x}=0.	
\end{equation}
\end{theorem}

\begin{proof}
Use Definition~\ref{D1}.
\end{proof}

\begin{theorem}\label{T3} 
The function $f(x)$ is of slow increase if and only if  $\frac{f(x)}{x^{\alpha}}$ has negative derivative (from a certain $x_{\alpha}$) for all $\alpha&gt;0$.
\end{theorem}

\begin{proof} 
We have
\begin{equation}\label{A3}
\frac{d}{dx}\left(\frac{f(x)}{x^{\alpha}}\right)=\frac{f(x)}{x^{\alpha+1}}\left(\frac{xf'(x)}{f(x)}-\alpha\right).
\end{equation}
Therefore if limit (\ref{A1}) holds we obtain that for all  $\alpha&gt;0$, 
\begin{equation}\label{A4}
\frac{d}{dx}\left(\frac{f(x)}{x^{\alpha}}\right)&lt; 0
\end{equation} 
for $x&gt; x_{\alpha}$. On the other hand, if (\ref{A4}) holds (for $x&gt;  x_{\alpha}$), (\ref{A3}) gives
\[0&lt;\frac{xf'(x)}{f(x)}&lt;\alpha.\]
Consequently we obtain (note that $\alpha$ is arbitrary)
\[\lim_{x\rightarrow \infty}\frac{x f'(x)}{f(x)}= 0.\]
That is, equation (\ref{A1}). 
\end{proof}

The following theorem justifies the term ``slow increase''. 

\begin{theorem}\label{T4} 
If the function $f(x)$ is of slow increase then
\begin{equation}\label{A5}
\lim_{x\rightarrow \infty}\frac{f(x)}{x^{\beta}} =0
\end{equation}
for all $\beta&gt;0$.
\end{theorem}

\begin{proof} 
Let $\alpha&gt;0$ be such that $\alpha&lt;\beta$. Then $\frac{f(x)}{x^{\alpha}}$ has a negative derivative (for $x&gt;x_{\alpha}$), then it is decreasing, therefore it is bounded $0&lt;
\frac{f(x)}{x^{\alpha}}&lt;M$. So
\[\lim_{x\rightarrow \infty}\frac {f(x)}{x^{\beta}} = \lim_{x\rightarrow \infty}\frac{f(x)}{x^{\alpha}}. \frac{1}{x^{\beta-\alpha}}= 0. \]
\end{proof}

\begin{corollary} 
If the function $f(x)$ is of slow increase then the following limits hold.
\begin{equation}\label{A6}
\lim_{x\rightarrow \infty}\frac {f(x)}{x} = 0,
\end{equation}
\begin{equation}\label{A7}
\lim_{x\rightarrow \infty}f'(x)=0.
\end{equation}
\end{corollary}

\begin{proof} 
Limit (\ref{A6}) is an immediate consequence of Theorem~\ref{T4}. Limit (\ref{A7}) is an immediate consequence of limit (\ref{A6}) and limit (\ref{A1}). 
\end{proof}

\begin{theorem} 
If the function $f(x)$ is of slow increase then 
\begin{equation}\label{A8}
\sum^{\infty}_{i=1}i^{\alpha}f(i)^{\beta}=\infty 
\end{equation}
for all $\alpha&gt; -1$ and for all $\beta$.
\end{theorem}

\begin{proof} 
We have
\begin{equation}\label{A9}
\sum^{\infty}_{i=1}i^{\alpha}f(i)^{\beta}= \sum^{\infty}_{i=1}\left(i^{\alpha+1}f(i)^{\beta}\right)\frac{1}{i}.
\end{equation}
Now, it is well-known that
\begin{equation}\label{A10}
\sum^{\infty}_{i=1}\frac{1}{i}=\infty.
\end{equation} 
On the other hand, we have (note that $\alpha +1&gt;0$)
\begin{equation}\label{A11}
\lim_{i\rightarrow \infty}i^{\alpha+1}f(i)^{\beta}=\infty.
\end{equation}
Limit (\ref{A11}) is clearly true if $\beta \geq 0$. If $\beta &lt; 0$ limit (\ref{A11}) is a direct consequence of Theorem~\ref{T2} ($f(x)^{-\beta}$ is of slow increase) and Theorem~\ref{T4}. 

Finally, equations (\ref{A9}), (\ref{A10}) and (\ref{A11}) give equation (\ref{A8}). 
\end{proof}

\begin{theorem} 
If the function $f(x)$ is of slow increase then  the following limit holds
\begin{equation}\label{A12}
\lim_{x\rightarrow \infty}\frac{\int^{x}_{a}t^{\alpha}f(t)^{\beta}\ dt}{\frac{x^{\alpha+1}}{\alpha+1}f(x)^{\beta}} =1
\end{equation}
for all $\alpha&gt; -1$ and for all $\beta$. 
\end{theorem}

\begin{proof} 
We have (see (\ref{A11}))
\[\lim_{x\rightarrow \infty}\frac{x^{\alpha+1}}{\alpha+1}f(x)^{\beta}=\infty.\]
On the other hand, the function $t^{\alpha}f(t)^{\beta}$ is either increasing or decreasing. 

Use Theorem~\ref{T2}  and Theorem~\ref{T3} in the case $\alpha&lt;0$, $\beta&gt;0$ and $\alpha&gt;0$, $\beta&lt;0$. The others cases are trivial.

Consequently (\ref{A8}) implies,
\[\lim_{x\rightarrow \infty}\int^{x}_{a}t^{\alpha}f(t)^{\beta}\ dt =\infty.\]
Now, limit (\ref{A12}) is a direct consequence of the L'Hospital's rule and limit (\ref{A1}). 
\end{proof}

Some particular cases of this theorem are the following: 

If $\alpha= 0$ we have 

\begin{equation}\label{A13}
\int^{x}_{a}f(t)^{\beta}\ dt \sim x f(x)^{\beta}. 
\end{equation}

If $\alpha= 0$ and  $\beta=1$ we have

\begin{equation}\label{A14}
\int^{x}_{a}f(t)\ dt \sim x f(x). 
\end{equation}

If $\alpha= 0$ and $\beta=-1$ we have

\begin{equation}\label{A15}
\int^{x}_{a}\frac{1}{f(t)}\ dt \sim \frac{x}{f(x)}. 
\end{equation}

\begin{theorem}\label{T8} 
If the function $f(x)$ is of slow increase and $C$ is a constant then the following limit holds
\begin{equation}\label{A16}
\lim_{x\rightarrow \infty}\frac{f(x+C)}{f(x)} =1.
\end{equation}
\end{theorem} 

\begin{proof} 
If $C&gt;0$, applying the Lagrange's Theorem we obtain 
\begin{equation}\label{A17}
0\leq \frac{f(x+C)-f(x)}{f(x)}= \frac{C f'(\xi)}{f(x)}, \qquad (x&lt;\xi&lt;x+C).
\end{equation}
Equations (\ref{A17}) and (\ref{A7}) give (\ref{A16}). In the same way can be proved the case  $C&lt;0$. 
\end{proof}

\begin{theorem}\label{T9} 
If the function $f(x)$ is of slow increase, $f'(x)$ is decreasing and $C &gt;0 $ then the following limit holds
\begin{equation}\label{A18}
\lim_{x\rightarrow \infty}\frac{f(C x)}{f(x)} =1.
\end{equation}
\end{theorem}
 
\begin{proof}
Suppose that $C&gt;1$. Applying Lagrange's theorem we obtain  
\begin{equation}\label{A19}
0\leq \frac{f(Cx)-f(x)}{f(x)}= \frac{(Cx-x) f'(\xi)}{f(x)}\leq (C-1)\frac{x f'(x)}{f(x)}, \quad (x&lt;\xi&lt;Cx).
\end{equation}
Equations (\ref{A19}) and (\ref{A1}) give (\ref{A18}).

Suppose that $C&lt;1$. Applying Lagrange's theorem we obtain
\begin{equation}\label{A20}
0\leq \frac{f(x)-f(Cx)}{f(Cx)}= \frac{(x-Cx) f'(\xi)}{f(Cx)}\leq \frac{1-C}{C}\frac{Cx f'(Cx)}{f(Cx)}, \quad (Cx&lt;\xi&lt;x).
\end{equation}
Equations (\ref{A20}) and (\ref{A1}) give (\ref{A18}). 
\end{proof} 

\begin{theorem}\label{T10}
If the function $f(x)$ is of slow increase, $f'(x)$ is decreasing and $0&lt; C_1\leq g(x)\leq C_2$ then the following limit holds.
\begin{equation}\label{A21}
\lim_{x\rightarrow \infty}\frac{f(g(x) x)}{f(x)} =1.
\end{equation}
\end{theorem} 

\begin{proof} 
We have
\begin{equation}\label{A22}
\frac{f(C_1 x)}{f(x)}\leq \frac{f(g(x)x)}{f(x)}\leq \frac{f(C_2 x)}{f(x)}.
\end{equation}
Equation (\ref{A22}) and Theorem~\ref{T9} give (\ref{A21}). 
\end{proof}





\section{Applications to Integer Sequences } 





In this section we consider only functions of slow increase that have decreasing derivative.

Let $A_n$ be a strictly increasing sequence of positive integers such that
\begin{equation}\label{A23}
A_n\sim n^s f(n),	\qquad (A_1&gt;1)
\end{equation}
and $f(x)$ is a function of slow increase. 

Let $\psi(x)$ be the number of  $A_n$ that do not exceed  $x$. 

\begin{example}\label{E11}
If $A_n=p_n$ is the sequence of prime numbers we have (Prime Number Theorem) $s=1$ and $f(x)= \log x$. If $A_n = c_{n,k}$ is the sequence of numbers with $k$ prime factors we have $s=1$ and $ f(x) = \frac{(k-1)!\log x}{(\log\log x)^{k-1}}$ (see \cite {Jakim2}). If $A_n=p_n^2$ we have $s=2$ and $f(x)= \log^2 x$.
\end{example}

\begin{remark} 
Note that:
(i) Theorem~\ref{T4} implies that $s\geq 1$ in equation (\ref{A23}).

(ii)  There  exists a strictly increasing sequence $A_n$ that satisfies (\ref{A23}), for example $A_n= \lfloor n^s f(n) \rfloor$.

(iii) If the function $g(x)$ is of slow increase then $\frac{g(A_n)}{g(n)}\rightarrow l\Leftrightarrow \frac{g(n^s f(n))}{g(n)}\rightarrow l$ and $\frac{g(A_n)}{g(n)}\rightarrow \infty \Leftrightarrow \frac{g(n^s f(n))}{g(n)}\rightarrow \infty$, because (Theorem~\ref{T10}) $g(A_n)\sim g(n^s f(n))$. 
\end{remark}

\begin{theorem} 
If $A_n$ satisfies (\ref{A23}) and $g(x)$ is a function of slow increase then the following equations hold
\begin{equation}\label{A24}
A_{n+1}\sim A_n, 
\end{equation}
\begin{equation}\label{A25}
\lim_{n\rightarrow \infty}\frac{A_{n+1}-A_n}{A_n} =0,
\end{equation}
\begin{equation}\label{A26}
\log A_{n+1}\sim \log A_n,
\end{equation}
\begin{equation}\label{A27}
g(A_{n+1})\sim g(A_n),
\end{equation}
\begin{equation}\label{A28}
\log A_n \sim s \log n,
\end{equation}
\begin{equation}\label{A29} 
\log \log A_n \sim \log \log n,
\end{equation}
\[\lim_{x\rightarrow \infty}\frac{\psi(x)}{x}=0.\]
\end{theorem}

\begin{proof}
Equation (\ref{A24}) is an immediate consequence of  equation (\ref{A23}) and Theorem~\ref{T8}. Equation (\ref{A25}) is an immediate consequence of equation (\ref{A24}). Equations (\ref{A26}) and (\ref{A27}) are an immediate consequence of equation (\ref{A24}) and Theorem~\ref{T10}. Equation (\ref{A28}) is an direct consequence of equations (\ref{A23}) and (\ref{A2}). Equation (\ref{A29}) is an direct consequence of (\ref{A28}). The last limit is an immediate consequence of (\ref{A23}) ($(A_n/n) \rightarrow \infty $) and  (\ref{A24}). 
\end{proof}

\begin{theorem}\label{T14}
If $A_n$ satisfies (\ref{A23}) and $g(x)$ is a function of slow increase then the following equation holds (note that $l\geq 1$).
\begin{equation}\label{A30}
g (A_n)\sim l g( n)	\Leftrightarrow g( \psi(x))\sim \frac{1}{l}g (x).
\end{equation}
In particular (see (\ref{A28}) and (\ref{A29}))
\begin{equation}\label{A31}
\log A_n\sim s \log n	\Leftrightarrow \log \psi(x)\sim \frac{1}{s}\log x,
\end{equation}
\begin{equation}\label{A32}
\log \log A_n\sim \log \log n	\Leftrightarrow \log \log \psi(x)\sim \log \log x.
\end{equation}
\end{theorem}

\begin{proof}
We have
\begin{eqnarray}
g( \psi(x))\sim \frac{1}{l}g( x) &amp;\Rightarrow&amp; g (\psi(A_n))\sim \frac{1}{l}g (A_n)	\Rightarrow g( n) \sim \frac{1}{l}g( A_n) \nonumber\\&amp;\Rightarrow&amp; g( A_n) \sim l g( n).
\nonumber
\end{eqnarray}
On the other hand 
\begin{eqnarray}\label{A33}
g( A_n )\sim l g( n) \Rightarrow \ g( A_n) \sim l g (\psi(A_n))\Rightarrow g( \psi(A_n))\sim \frac{1}{l}g( A_n).
\end{eqnarray}
If $ A_n\leq x&lt; A_{n+1}$ we have
\[\frac{g( \psi(A_n))}{\frac{1}{l}g( A_{n+1})}\leq \frac{g( \psi(x))}{\frac{1}{l}g( x)}
\leq \frac{g( \psi(A_n))}{\frac{1}{l}g( A_n)}.\]
Now, both sides have limit 1 (see (\ref{A33}) and (\ref{A27})).  
\end{proof}

We shall need the following well-known lemma (see \cite[p.\ 332]{ReyPastor5}).

\begin{lemma}\label{L15}
Let $\sum^{\infty}_{i=1}a_i$ and $\sum^{\infty}_{i=1}b_i$ be two series of positive terms such that $\lim_{i\rightarrow \infty}\frac{a_i}{b_i}=1$. Then if $\sum^{\infty}_{i=1}b_i$ is divergent, the following limit holds.
\[\lim_{n\rightarrow \infty}\frac{\sum^{n}_{i=1}a_i}{\sum^{n}_{i=1}b_i}=1.\]
\end{lemma}

In the following theorem we shall obtain  information on $\psi(x)$ when $s=1$ (see (\ref{A23})) and $f(A_n)\sim f(n)$.

\begin{theorem} \label{T16}
If $f(A_n)\sim f(n)$ then 
\begin{equation}\label{A34}
A_n\sim  n f(n)	\Leftrightarrow \psi(x)\sim \frac{x}{f(x)}\Leftrightarrow \psi(x)\sim  \int^{x}_{a}\frac{1}{f(t)}\ dt \Leftrightarrow \sum_{A_i\leq x}f(A_i)\sim x.
\end{equation}
Besides if $g(x)$ is a function of slow increase and $g(A_n)\sim l'g(n)$ then
\begin{equation}\label{A35}
\psi(x)\sim \frac{\sum_{A_i\leq x}g(A_i)^{\beta}}{g(x)^{\beta}}
\end{equation}
for all $\beta$.
\end{theorem}	

\begin{proof} 
We have (note that $\frac{x}{f(x)}\rightarrow \infty$, see (\ref{A6}))
\begin{eqnarray}
\psi(x)\sim \frac{x}{f(x)}&amp;\Rightarrow&amp; \psi(A_n)\sim \frac{A_n}{f(A_n)}\Rightarrow n \sim 	
\frac{A_n}{f(A_n)}\Rightarrow A_n \sim n f(A_n)\nonumber\\&amp;\Rightarrow&amp; A_n \sim n f(n).\nonumber
\end{eqnarray}  
On the other hand
\begin{eqnarray}\label{A36}
A_n \sim n f(n)&amp;\Rightarrow&amp; A_n \sim \psi(A_n) f(n)\Rightarrow \psi(A_n)\sim \frac{A_n}{f(n)}
\nonumber\\&amp;\Rightarrow&amp; \psi(A_n)\sim \frac{A_n}{f(A_n)}.	
\end{eqnarray}
If $ A_n\leq x&lt; A_{n+1}$ we have (note that $\frac{x}{f(x)}$ is increasing, see Theorem~\ref{T3})
\begin{equation}\label{A37}
\frac{\psi(A_n)}{\frac{A_{n+1}}{f(A_{n+1})}}\leq \frac{\psi(x)}{\frac{x}{f(x)}}\leq 
\frac{\psi(A_n)}{\frac{A_n}{f(A_n)}}.
\end{equation}
Now, both sides have limit 1 (see (\ref{A36}), (\ref{A24}) and (\ref{A27})). Consequently
\[A_n\sim  n f(n)	\Rightarrow \psi(x)\sim \frac{x}{f(x)}.\]
On the other hand (see (\ref{A15}))
\[\psi(x)\sim \frac{x}{f(x)}\Leftrightarrow \psi(x)\sim  \int^{x}_{a}\frac{1}{f(t)}\ dt.\]
Note that (see (\ref{A13}))
\[\int^{n}_{a}g(x)^{\beta}\ dx \sim n g(n)^{\beta}.\]
Therefore as $g(x)^{\beta}$ is either increasing or decreasing,
\begin{equation}\label{A38}
\sum^{n}_{i=1}g(i)^{\beta}=\int^{n}_{a}g(x)^{\beta}\ dx +h(n)\sim n g(n)^{\beta}.
\end{equation}
Equation (\ref{A38}), $g(A_n)^{\beta}\sim l'^{\beta} g(n)^{\beta}$ and Lemma~\ref{L15} give
\[\sum^{n}_{i=1}g(A_i)^{\beta}\sim n l'^{\beta} g(n)^{\beta}.\]
That is
\[\sum_{A_i\leq A_n}g(A_i)^{\beta}\sim \psi(A_n) g(A_n)^{\beta}.\]
Consequently 
\begin{equation} \label{A39}
\psi(A_n)\sim \frac{\sum_{A_i\leq A_n}g(A_i)^{\beta}}{g(A_n)^{\beta}}.
\end{equation}
If $ A_n\leq x &lt; A_{n+1}$ we have $(\beta&gt;0)$
\[\frac{\psi(A_n)}{\frac{\sum_{A_i\leq A_n}g(A_i)^{\beta}}{g(A_n)^{\beta}}}\leq
\frac{\psi(x)}{\frac{\sum_{A_i\leq x}g(A_i)^{\beta}}{g(x)^{\beta}}}\leq \frac{\psi(A_n)}{\frac{\sum_{A_i\leq A_n}g(A_i)^{\beta}}{g(A_{n+1})^{\beta}}}.\]
Now, both sides have limit 1 (see (\ref{A39}) and (\ref{A27})). Therefore
\[\psi(x)\sim \frac{\sum_{A_i\leq x}g(A_i)^{\beta}}{g(x)^{\beta}}.\]
That is, equation (\ref{A35}). If $\beta&lt;0$, the proof of (\ref{A35}) is the same.

Consequently if $g(x)=f(x)$ and $\beta = 1$ we find that
\[\psi(x)\sim \frac{x}{f(x)}\Leftrightarrow  \sum_{A_i\leq x}f(A_i)\sim x.\]
\end{proof} 
 
\begin{example} 
Let us consider the sequence $p_n$ of prime numbers, in this case we have (Prime Number Theorem) $p_n\sim n \log n$ and $\psi(x)=\pi(x)\sim x/(\log x)$. Let us consider the sequence  $c_{n,k}$ of numbers with $k$ prime factors, in this case we have $ c_{n,k} \sim \frac{(k-1)!n \log n}{(\log\log n)^{k-1}}$ (see Example~\ref{E11}) and (Landau's Theorem)   (see \cite {Hardy1, Jakim2}) $\psi(x)\sim \frac{x (\log\log x)^{k-1}}{(k-1)! \log x}$. 
\end{example}

In the following general theorem we obtain  information on $\psi(x)$ if $f(A_n)\sim l f(n)$. 

Theorem~\ref{T16} is a particular case of this Theorem.

\begin{theorem}\label{T18} 
If $f(A_n)\sim l f(n)$ then 
\begin{eqnarray}
A_n\sim  n^s f(n)	&amp;\Leftrightarrow&amp; \psi(x)\sim l^{\frac{1}{s}}\frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}}\Leftrightarrow \psi(x)\sim \frac{l^{\frac{1}{s}}}{s}\int^{x}_{a}\frac{t^{-1+\frac{1}{s}}}{f(t)^{\frac{1}{s}}}\ dt\nonumber\\&amp;\Leftrightarrow&amp; \sum_{A_i\leq x}f(A_i)^{\frac{1}{s}}\sim l^{\frac{1}{s}} x^{\frac{1}{s}}.\nonumber
\end{eqnarray}
Besides if $g(x)$ is a function of slow increase and $g(A_n)\sim l'g(n)$ then
\begin{equation}\label{A40}
\psi(x)\sim \frac{\sum_{A_i\leq x}g(A_i)^{\beta}}{g(x)^{\beta}}
\end{equation}
for all $\beta$.
\end{theorem}

\begin{proof} 
The proof that
\[A_n\sim  n^s f(n)	\Leftrightarrow \psi(x)\sim l^{\frac{1}{s}}\frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}}\]
is the same as in Theorem~\ref{T16}.
Now, see equation (\ref{A12}),
\[\int^{x}_{a}\frac{t^{-1+\frac{1}{s}}}{s f(t)^{\frac{1}{s}}}\ dt \sim \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}}.\]
Therefore
\[\psi(x)\sim l^{\frac{1}{s}}\frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}}\Leftrightarrow\psi(x)\sim \frac{l^{\frac{1}{s}}}{s}\int^{x}_{a}\frac{t^{-1+\frac{1}{s}}}{f(t)^{\frac{1}{s}}}\ dt. \]
The proof of the equation (\ref{A40}) is the same as in Theorem~\ref{T16}. If $g(x)=f(x)$ and $\beta = 1/s$ then we find that
\[\psi(x)\sim l^{\frac{1}{s}}\frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}} \Leftrightarrow \sum_{A_i\leq x}f(A_i)^{\frac{1}{s}}\sim l^{\frac{1}{s}} x^{\frac{1}{s}}.\]
\end{proof}


\begin{example} 
Let us consider the following sequence of positive integers (see Theorem~\ref{T22})
\[A_n=\sum^{n}_{i=1}p_i^k \sim \frac{n^{k+1}}{k+1}\log^k n\]
where $k$ is a positive integer. In this case we have $s=k+1$, $f(x)= \frac{\log^k  x}{k+1}$ and $l= (k+1)^k$. Consequently
\[\psi(x)\sim (k+1)\frac{x^{\frac{1}{k+1}}}{\left(\log x\right)^{\frac{k}{k+1}}}.\]
\end{example}

Let us consider the sequence $P_n$ of the $A_n$ powers. For example, if $A_n=p_n$ is the sequence of prime numbers, $P_n$ is the sequence of prime powers . Let $\lambda(x)$ be the number of  $P_n$ that do not  exceed $x$.  

\begin{theorem} 
If $A_n$ satisfies (\ref{A23}) then
\begin{equation}\label{A41}
\lambda(x)\sim \psi(x).
\end{equation} 
\end{theorem} 

\begin{proof}
The $A_i\leq x$ are $A_1, A_2,\ldots, A_{\psi(x)}$. Let us write
\[A_i^{\alpha_i}= x, \qquad (i=1,2,\ldots, \psi(x)).\]
Therefore
\[\alpha_i = \frac{\log x}{\log A_i},\qquad(i=1,2,\ldots, \psi(x)).\]

We have the following inequalities
\begin{equation}\label{A42}
\psi(x)\leq \lambda(x)\leq \sum^{\psi(x)}_{i=1}[\alpha_i]\leq \sum^{\psi(x)}_{i=1}\alpha_i
= \log x \sum^{\psi(x)}_{i=1}\frac{1}{\log A_i}.
\end{equation}
Equation (\ref{A28}) gives 
\begin{equation}\label{A43}
\frac{1}{\log A_n}\sim \frac{1}{s \log n}.
\end{equation}
Note that (see (\ref{A15}))
\[\int^{x}_{2}\frac{1}{\log t}\ dt \sim \frac{x}{\log x}.\]
Now,   
\begin{eqnarray}\label{A44}
&amp;&amp;\frac{1}{\log A_1}+\sum^{\psi(x)}_{i=2}\frac{1}{s\log i}=\frac{1}{\log A_1}+\frac{1}{s}\sum^{\psi(x)}_{i=2}\frac{1}{\log i}\nonumber\\&amp;=&amp;
\frac{1}{s}\int^{\psi(x)}_{2}\frac{1}{\log t}\ dt +O(1) \sim \frac{\psi(x)}{s \log \psi(x)}.
\end{eqnarray}
Equations (\ref{A43}), (\ref{A44}) and Lemma~\ref{L15} give
\begin{equation}\label{A45}
\sum^{\psi(x)}_{i=1}\frac{1}{\log A_i}\sim \frac{\psi(x)}{s \log \psi(x)}.
\end{equation}
Equations (\ref{A42}) and (\ref{A45}) give
\[\psi(x)\leq \lambda(x)\leq h(x) \frac{\psi(x)\log x}{s \log \psi(x)},\]
where $h(x)\rightarrow 1$. That is
\begin{equation}\label{A46}
1\leq \frac{\lambda(x)}{\psi(x)}\leq h(x) \frac{\log x}{s \log \psi(x)}.
\end{equation}
Finally, equations (\ref{A31}) and (\ref{A46}) give (\ref{A41}).  
\end{proof}

\begin{corollary} 
The following limit holds.
\[\lim_{x\rightarrow \infty}\frac{\sum^{\psi(x)}_{i=1}\left(\alpha_i-[\alpha_i]\right)}{\psi(x)}=0.\]
That is, the mean fractional part has limit zero.
\end{corollary}

\begin{theorem}\label{T22}  
If $A_n$ satisfies (\ref{A23}) then the following asymptotic formulas hold
\begin{equation}\label{A47}
\sum^{n}_{i=1}A_i^{\alpha} \sim \frac{n^{s\alpha +1}f(n)^{\alpha}}{s\alpha +1}\sim \frac{n}{s \alpha +1}\ A_n^{\alpha},\qquad (\alpha &gt;0),
\end{equation}
\begin{equation}\label{A48}
\sum_{A_i\leq x}A_i^{\alpha} \sim \frac{\psi (x)}{s \alpha +1}\ x^{\alpha},\qquad (\alpha &gt;0).
\end{equation}
\end{theorem}

\begin{proof} 
Let us consider the sum
\begin{equation}\label{A49}
1+2+\cdots+(n'-1)+\sum^{n}_{i=n'}\left(i^s f(i)\right)^{\alpha},  
\end{equation}
where $n'$ is a positive integer on interval $[a, \infty)$. Note that (see (\ref{A23}))
\begin{equation}\label{A50}
A_i^{\alpha}\sim \left(i^s f(i)\right)^{\alpha}.
\end{equation}
Note that the function $x^s f(x)$ is increasing and therefore we have
\begin{equation}\label{A51}
\sum^{n}_{i=n'}\left(i^s f(i)\right)^{\alpha}=\int^{n}_{n'}x^{s\alpha}f(x)^{\alpha}\ dx+O\left(n^{s\alpha}f(n)^{\alpha}\right).
\end{equation}
On the other hand (see (\ref{A12}))
\begin{equation}\label{A52}
\int^{n}_{n'}x^{s\alpha}f(x)^{\alpha}\ dx \sim \frac{n^{s\alpha +1}f(n)^{\alpha}}{s\alpha +1}.
\end{equation}
Equations (\ref{A49}), (\ref{A51}) and (\ref{A52}) give  
\begin{equation}\label{A53}
1+2+\cdots+(n'-1)+\sum^{n}_{i=n'}\left(i^s f(i)\right)^{\alpha}\sim \frac{n^{s\alpha +1}f(n)^{\alpha}}{s\alpha +1}\sim \frac{n}{s \alpha +1}A^{\alpha}_{n}.
\end{equation}
Finally,  (\ref{A53}), (\ref{A50}) and Lemma~\ref{L15} give (\ref{A47}). 

If we substitute $n=\psi (A_n)$ into equation (\ref{A47}) and proceed as in Theorem~\ref{T14} and Theorem~\ref{T16} then we obtain (\ref{A48}).  
\end{proof}

\begin{remark} 
Equations (\ref{A47}) and (\ref{A48}) when $A_n=p_n$ is the sequence of prime numbers were obtained by S\'alat and Zn\'am \cite {SalatZnam6}, more precise formulas when  $\alpha$ is a positive integer were obtained by Jakimczuk \cite {Jakim3}. Equations (\ref{A47}) and (\ref{A48}) when $A_n=c_{n,k}$ is the sequence of  numbers with  $k$ prime factors were obtained by Jakimczuk \cite {Jakim2}.
\end{remark} 

Jakimczuk \cite {Jakim4} proved the following theorem. 

\begin{theorem} 
If $A_n$ satisfies (\ref{A23}) then the following formulas hold
\[\sum^{n}_{i=1}\log A_i = s\  n \log n-s\  n + n \log f(n) +o(n),\]
\[\lim_{n\rightarrow\infty}\frac{\sqrt[n]{A_1 A_2\ldots A_n}}{A_n}=\frac{1}{e^s}.\]
\end{theorem}

\begin{proof}
See \cite {Jakim4}. In that proof we supposed that 
\[\lim_{x\rightarrow \infty}\int^{x}_{a}\frac{t f'(t)}{f(t)}\ dt = \infty.\]
Consequently (L'Hospital's rule)
\begin{equation}\label{A54}
\lim_{x\rightarrow \infty}\frac{\int^{x}_{a}\frac{t f'(t)}{f(t)}\ dt}{x} = 0.
\end{equation}
This supposition is unnecessary since if the integral converges then (\ref{A54}) also holds.
\end{proof}

\begin{definition} 
The function of slow increase $f(x)$ is a {\em universal function} if
and only if for all sequence $A_n$ that satisfies (\ref{A23}) we have
$f(A_n)\sim l f(n)$ where $l$ depends of the sequence $A_n$.
\end{definition}

\begin{example} Equation (\ref{A28}) implies that $f(x)=\log x$ is an
universal function, in this case $l=s$. Equation (\ref{A29}) implies
that $f(x)=\log \log x$ is an universal function, in this case $l=1$
does not depend of the sequence $A_n$.
\end{example}

\begin{remark} 
Note that if $f(x)$ and $g(x)$ are universal functions then $f(x)^{\alpha}$ $(\alpha&gt;0)$, $Cf(x)$ $(C&gt;0)$ and $f(x)g(x)$ are  universal functions.
If $f(x)/g(x)$ is a function of slow increase then is  an universal function.
\end{remark}

\begin{theorem}  If $f(x)$ is an universal function and $A_n$ satisfies (\ref{A23}) then we have 
\[\psi(x)\sim \frac{\sum_{A_i\leq x}f(A_i)^{\beta}}{f(x)^{\beta}}\]
for all $\beta$. 
\end{theorem}

\begin{proof}
The proof is the same as in Theorem~\ref{T16} and Theorem~\ref{T18}.
\end{proof}

\begin{example} 
Since $f(x)=\log x $ is an universal function,
we have for all sequences $A_n$ satisfying (\ref{A23}) that
\[\psi(x)\sim \frac{\sum_{A_i\leq x}\log^{\beta} A_i}{\log^{\beta} x}.\]
In particular, if $\beta =1$ we have
\[\psi(x)\sim \frac{\sum_{A_i\leq x}\log A_i}{\log x}.\]
\end{example}

\begin{theorem} 
There exist functions of slow increase that are not universal functions. 
\end{theorem}  

\begin{proof}
We shall prove that the following function of slow increase
\[g(x)=e^{\frac{\log x}{\log\log x}},\]
is not an universal function. We shall prove that there exists a sequence $A_n$ that satisfies (\ref{A23}) and
\[\lim_{n\rightarrow \infty}\frac{g(A_n)}{g(n)}= \infty.\]
Since $A_n$ satisfies (\ref{A23}) we can write
\[A_n=h_1(n) n^s f(n),\]
where $h_1(n)\rightarrow 1$. Therefore
\begin{equation}\label{A55}
\frac{g(A_n)}{g(n)}=\exp\left(\frac{\log h_1(n)+s \log n+\log f(n)}{\log\log n+\log s+\log \left(1+\frac{\log f(n)}{s\log n}+\frac{\log h_1(n)}{s\log n} \right)}-\frac{\log n}{\log\log n}\right).	
\end{equation}
If $s&gt;1$ (\ref{A55}) becomes (see (\ref{A2}))
\[\frac{g(A_n)}{g(n)}=\exp\left(h_2(n)\frac{s \log n}{\log\log n}-\frac{\log n}{\log\log n}\right),\]
where $h_2(n)\rightarrow 1$. That is
\[\frac{g(A_n)}{g(n)}=\exp\left(h_3(n)\frac{(s-1) \log n}{\log\log n}\right),\]
where $h_3(n)\rightarrow 1$. Consequently we have
\[\lim_{n\rightarrow \infty}\frac{g(A_n)}{g(n)}= \infty.\]
This proves the theorem. In particular this limit is true if $f(x)=g(x)$.

To complete, we shall examine  the case $s=1$. In this case (\ref{A55}) becomes (note that  $\lim_{x\rightarrow 0}\frac{\log(1+x)}{x}=1$)  
\begin{eqnarray} 
\frac{g(A_n)}{g(n)}&amp;=&amp;\exp\left(\frac{\log h_1(n)+ \log n+\log f(n)}{\log\log n+h_4(n)\frac{\log f(n)}{\log n}+h_4(n)\frac{\log h_1(n)}{\log n} }-\frac{\log n}{\log\log n}\right)\nonumber\\&amp;=&amp; \exp\left(\frac{ \log n+\log f(n)}{\log\log n+h_4(n)\frac{\log f(n)}{\log n}+h_4(n)\frac{\log h_1(n)}{\log n} }-\frac{\log n}{\log\log n}+o(1)\right)\nonumber\\&amp;=&amp; \exp\left(\frac{ \log\log n\log f(n)-h_4(n)\log f(n)-h_4(n)\log h_1(n)}{(\log\log n)^{2} +h_4(n)\frac{\log\log n\log f(n)}{\log n}+h_4(n)\frac{\log\log n\log h_1(n)}{\log n} }+o(1)\right)\nonumber\\&amp;=&amp;\exp\left(h_5(n)\frac{\log f(n)}{\log\log n}+o(1)\right), \nonumber	
\end{eqnarray}
where $h_4(n)\rightarrow 1$ and $h_5(n)\rightarrow 1$. 
  
For example, if  $f(x)=g(x)$ then $\lim_{n\rightarrow \infty}\frac{g(A_n)}{g(n)}= \infty$.
If  $f(x)=\log^{\alpha}x$ $(\alpha&gt;0)$ then $\lim_{n\rightarrow \infty}\frac{g(A_n)}{g(n)}= e^{\alpha}$. If  $f(x)=\log\log x$ then $\lim_{n\rightarrow \infty}\frac{g(A_n)}{g(n)}= 1$.
\end{proof}
 

\section{Acknowledgements}

The author would like to thank the anonymous referees for their
valuable comments and suggestions for improving the original version of
this manuscript.

\begin{thebibliography}{99}
 
\bibitem{Hardy1}
G. H. Hardy and E. M. Wright, \textit{An Introduction to the Theory of Numbers}, Fourth Edition, 1960.

\bibitem{Jakim2}
R. Jakimczuk, A note on sums of powers which have a fixed number of prime factors, \textit{J.  Inequal.  Pure  Appl. Math.} \textbf{6} (2005), Article 31.

\bibitem{Jakim3}  
R. Jakimczuk, Desigualdades y f\'ormulas asint\'oticas para sumas de
potencias de primos, \textit{Bol. Soc. Mat. Mexicana} (3) \textbf{11}
(2005), 5--10.

\bibitem{Jakim4}
R. Jakimczuk, The ratio between the average factor in a product and the last factor, \textit{Mathematical Sciences: Quarterly Journal} \textbf{1} (2007), 53--62.

\bibitem{ReyPastor5}
J. Rey Pastor, P. Pi Calleja, C. Trejo, \textit{An\'alisis
Matem\'atico,} Volume I, Editorial Kapeluz, Octava Edici\'on,  1969.

\bibitem{SalatZnam6}
T. S\'alat and S. Zn\'am, On the sums of prime powers, \textit{Acta.
Fac. Rer. Nat. Univ. Com. Math.} \textbf{21} (1968), 21--25.

\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B99; Secondary 11N45.

\noindent \emph{Keywords: } 
Functions of slow increase, integer sequences, asymptotic formulas.

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received  September 14 2009;
revised version received  December 21 2009.
Published in {\it Journal of Integer Sequences}, December 23 2009.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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