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\begin{center}
\vskip 1cm{\LARGE\bf  \ The Riemann Zeta Function and the Fractional Part of Rational Powers\
\vskip 1cm}
\large
Tal Barnea\\
United Kingdom\\
\href{mailto: tal.q.barnea@gmail.com}{\tt tal.q.barnea@gmail.com} \\
\end{center}

\vskip .2 in

\begin{abstract}
We use elementary methods to find an explicit formula for the second term
of the continued fraction of the Riemann zeta function over positive
integers, with only finitely many exceptions. As a corollary to this,
we find connections between the Riemann zeta function and the fractional
parts of rational powers.
\end{abstract}

\section{Introduction}
Given an irrational real number $x$ we recall that the \textit{simple continued fraction} of it is
$$x=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}$$  where all $a_i$ are integers and $a_i  \ge  1$ for $i \ge 1$. Thus, the first term $a_0=\lfloor x  \rfloor$ and the second term is $a_1=\left\lfloor \frac{1}{x-a_0 }  \right\rfloor$. 

The Riemann zeta function is defined for real $s&gt;1$ as $$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}.$$
For all positive integers $n \ge 2$ we have that $1 &lt; \zeta(n)&lt;2$.
Therefore, the first term of $\zeta(n)$ in its simple continued fraction is always $1$. 

Let $a(n)$ denote the second term in the continued fraction of
$\zeta(n)$. According to Adams-Watters
in the description of sequence \seqnum{A013697} in the {\it On-Line
Encyclopedia of Integer Sequences} (OEIS) \cite{OEIS}, 
``It appears that $a(n) = 2^n - \left\lfloor \left(
\frac{4}{3} \right)^n \right\rfloor - k$, where $k$ is usually $2$,
but is sometimes $1$. Up to $n=1000$, the only values of $n$ where
$k = 1$ are $4$, $5$, $13$, $14$, and $17$. That is, $$\left\lfloor
\frac{1}{\zeta(n)-1} \right\rfloor=2^n-\left\lfloor \left( \frac{4}{3}
\right)^n \right\rfloor-k,$$ where $k=1$ or $k=2$''.  We first prove
that this formula holds for all $n \geq 2$.

\begin{theorem}  \label{theorem:Main}
The equality $$\left\lfloor \frac{1}{\zeta(n)-1} \right\rfloor=2^n-\left\lfloor \left( \frac{4}{3} \right)^n \right\rfloor-2,$$  holds for all natural numbers, $n \ge 2$, with finitely many exceptional cases. In the exceptional cases we instead have $$\left\lfloor \frac{1}{\zeta(n)-1} \right\rfloor=2^n-\left\lfloor \left( \frac{4}{3} \right)^n \right\rfloor-1.$$
\end{theorem}

We write $ \{ x \}=x-\lfloor x \rfloor$ , for the fractional part of $x$. Given coprime integers  $p&gt;q&gt;1$, we consider the set $\left\{ \left\{ \left( \frac{p}{q} \right)^{n} \right\} \mid n \in \N \right\}$. Vijayaraghavan \cite{Vi} proved that this set has infinitely many limit points, but otherwise not much is known about its distribution.  For real $s$ write $\varepsilon_x(s)=(2x)^s \left( \left( \frac{2}{3} \right)^s + \left( \frac{1}{2} \right)^s \right)^2$ and for simplicity we write $\varepsilon(s)=\varepsilon_1(s)$. As a corollary of Theorem~\ref{theorem:Main} we obtain the following result. 

\begin{corollary} \label{corollary:FractionalParts} 
For all natural numbers, $n \ge 2$, with finitely many exceptional cases, we have that
$$1&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;1+\varepsilon(n).$$
In the exceptional cases we instead have
$$0&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;\varepsilon(n).$$
\end{corollary}
Corollary~\ref{corollary:FractionalParts} implies the set $\left\{  \left\{ \frac{1}{\zeta(n)-1}  \right\}  \mid n \in \N \right\}$ has infinitely many limit points.

\begin{corollary} \label{corollary:Egypt}
For a natural number, $n \ge2$, $\zeta(n)$ can only be of the form $1+ \frac{1}{m}$, where $m$ is a natural number, finitely many times.
\end{corollary}

\begin{conjecture} \label{conjecture1}
The finite exceptions of Theorem~\ref{theorem:Main} only occur when $n=4,5,13,14,17$.
\end{conjecture} 
Using Mathematica we checked when $\left\{ \left( \frac{4}{3} \right)^n \right\}$ is less than $10^{-9}$ for $n \le 5,000,000$, and no examples were found. As $\varepsilon(n)&lt;10^{-9}$ for $n \ge 176$, Corollary~\ref{corollary:FractionalParts} implies that the conjecture holds for $176 \leq n \leq 5,000,000$. Since Adams-Watters verified the conjecture for $n \leq 1000$, we conclude the conjecture holds for $n \le 5,000,000$.

Using similar methods to Theorem~\ref{theorem:Main} and Corollary~\ref{corollary:FractionalParts}, results can also be obtained for other functions of the form $$\sum_{i=1}^{\infty} \frac{\left(-1\right)^{t_i}}{{u_i}^s},$$ where $t_i$ and $u_i$ are some sequence. The Riemann zeta function is such that $t_i=0$ and $u_i=i$. For example for the Dirichlet beta function, $\beta(s)$, where $t_i=i-1$ and $u_i=2i-1$, we can show
$$\left\{ \left\{\frac{1}{1-\beta(n)} \right\}-\left\{\left(\frac{9}{5} \right)^n\right\}+\left\{\left(\frac{9}{7} \right)^n\right\}-\left\{\left(\frac{27}{25} \right)^n\right\} \right\} = 1-\left(\frac{9}{11} \right)^n+\mathcal{O}\left( \left(\frac{27}{35} \right)^n \right).$$
For $P(s)$ where $t_i=0$ and $u_i$ is the $i^{th}$ prime, we can show that
$$1-\left(\frac{8}{9} \right)^s - 2\left( \frac{2}{3} \right)^s - \frac{1}{2^s}&lt; \frac{1}{P(s)} - \frac{1}{\zeta(s)-1} &lt; 1+\left(\frac{8}{9} \right)^s+2\left( \frac{8}{15} \right)^s + \left( \frac{8}{25} \right)^s.$$

The most interesting of these results using different functions is given in Theorem~\ref{theorem:Zeta}. This is using the function for when $t_i=0$ and $u_i= \frac{2i}{3}$.

\begin{theorem} \label{theorem:Zeta}
For $n$ large enough we have
$$1-\left( \frac{8}{9} \right)^n- \left( \frac{2}{3} \right)^n &lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{  \frac{\left( \frac{2}{3} \right)^n}{\zeta(n)-1} \right\} &lt; $$
$$1-\left( \frac{8}{9}\right)^n- \left( \frac{2}{3} \right)^n +\varepsilon(n) + \varepsilon_{\frac{2}{3}}(n).$$
\end{theorem}


\section{Analytic Results}
\begin{proposition} \label{proposition:Lower}
For all $n \ge 2$ we have that $$1&lt;\frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2.$$
\end{proposition}
\begin{proof}
For $n=2$ and $n=3$, it can be checked that the inequality is true.
For $n \ge 4$ we prove the proposition by contradiction. Assume $$1\geq \frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2.$$
We will first cancel out some terms from our inequality.
Our assumption implies that  $$2^n-\left( \frac{4}{3} \right)^n-1 \geq \frac{1}{\zeta(n)-1},$$
$$\frac{6^n-4^n-3^n}{3^n} \geq \frac{1}{\zeta(n)-1},$$
$$\frac{3^n}{6^n-4^n-3^n} \leq \zeta(n)-1,$$
$$3^n \leq \left(6^n-4^n-3^n \right) \sum_{i=2}^{\infty} \frac{1}{i^n},$$
\begin{equation}
\label{eq:1}
3^n+ \sum_{i=2}^{\infty} \frac{3^n}{i^n} + \sum_{i=2}^{\infty} \frac{4^n}{i^n} \leq \sum_{i=2}^{\infty} \frac{6^n}{i^n}.
\end{equation}
On the other hand, the following three inequalities are obvious:
\begin{equation}
\label{eq:2}
\sum_{i=2}^{\infty} \frac{3^n}{i^n} &gt; \left(\frac{3}{2} \right)^n +1,
\end{equation}
\begin{equation}
\label{eq:3}
\sum_{i=2}^{\infty} \frac{4^n}{i^n} &gt; 2^n+ \left(\frac{4}{3} \right)^n+1,
\end{equation}
\begin{equation}
\label{eq:4}
 \sum_{i=2}^{\infty} \frac{6^n}{i^n} &lt; 3^n+2^n+\left( \frac{3}{2} \right)^n+ \left( \frac{6}{5} \right)^n+1+\int_{6}^{\infty} \left(\frac{6}{t} \right)^ndt=3^n+2^n+\left( \frac{3}{2} \right)^n+ \left( \frac{6}{5} \right)^n+1+\frac{6}{n-1}.
\end{equation}
Thus, substituting \eqref{eq:2}, \eqref{eq:3}, and \eqref{eq:4} into \eqref{eq:1}, we have
$$3^n+\left(\frac{3}{2}\right)^n +1+2^n+\left(\frac{4}{3} \right)^n +1 &lt; 3^n+2^n+\left(\frac{3}{2} \right)^n + \left(\frac{6}{5} \right)^n +1+\frac{6}{n-1},$$
$$\left(\frac{4}{3} \right)^n +1 &lt; \left(\frac{6}{5} \right)^n + \frac{6}{n-1}.$$
It is clear that the above statement is false for $n \ge 7$. Specifically it can be checked that it is false for $n \ge 4$. This is a contradiction and thus, $$1&lt; \frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2,$$ for all $n \ge 2$.
\end{proof}

\begin{proposition} \label{proposition:Upper}
For $n \ge 2$ we have that $$\frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2 &lt; 1+ \varepsilon(n).$$
\end{proposition}
\begin{proof}
Consider
$$\frac{1}{\zeta(n)-1}=\frac{1}{\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\cdots} &lt; \frac{1}{\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}}.$$
Thus,
$$\frac{1}{\zeta(n)-1} &lt; \frac{2^n}{1+\frac{2^n}{3^n} +\frac{1}{2^n}}=$$
$$2^n \left(1-\left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right) +\left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right)^2 - \left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right)^3 + \cdots \right)&lt;$$
$$2^n \left(1-\left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right) +\left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right)^2 \right)=2^n -\left( \frac{4}{3} \right)^n -1+2^n \left( \frac{2^n}{3^n}+ \frac{1}{2^n} \right)^2.$$
We conclude that $$\frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2 &lt; 1+ \varepsilon(n),$$ for $n \ge 2$.
\end{proof}


\section{Proof of Results}
The following lemma is obvious.
\begin{lemma} \label{lemma:Floor}
If $$A&lt;x&lt;B,$$ then $$A-1&lt;\lfloor{x}\rfloor&lt;B.$$
\end{lemma}

\begin{proposition} \label{proposition:Floor}
For all natural numbers $n \ge 2$, we have that $$\left\lfloor \frac{1}{\zeta(n)-1} \right\rfloor=2^n-\left\lfloor \left( \frac{4}{3} \right)^n \right\rfloor-k,$$    where $k=1$ or $k=2$.
\end{proposition}
\begin{proof}
It is easy to check the equality holds for $2 \le n \le 4$.
We know from Proposition~\ref{proposition:Lower} and Proposition~\ref{proposition:Upper}  that $$1&lt;\frac{1}{\zeta(n)-1}-2^n+\left( \frac{4}{3} \right)^n+2&lt;1+\varepsilon(n).$$
Applying Lemma~\ref{lemma:Floor}  twice to this inequality we obtain that $$-1&lt;\left\lfloor\frac{1}{\zeta(n)-1}\right\rfloor - 2^n +\left\lfloor\left( \frac{4}{3} \right)^n\right\rfloor+2&lt;1+\varepsilon(n).$$
 As $\lim\limits_{n\to\infty} \varepsilon(n)=0$, for $n \ge 5$  we have that $$-1&lt;\left\lfloor\frac{1}{\zeta(n)-1}\right\rfloor - 2^n +\left\lfloor\left( \frac{4}{3} \right)^n\right\rfloor+2&lt;2.$$
It is clear that $\left\lfloor\frac{1}{\zeta(n)-1}\right\rfloor - 2^n +\left\lfloor\left( \frac{4}{3} \right)^n\right\rfloor+2$ is an integer for $n \ge 2$.  Thus, $$\left\lfloor\frac{1}{\zeta(n)-1}\right\rfloor - 2^n +\left\lfloor\left( \frac{4}{3} \right)^n\right\rfloor+2=0 \textrm{ or } 1.$$  Hence, $$\left\lfloor\frac{1}{\zeta(n)-1}\right\rfloor=2^n -\left\lfloor\left( \frac{4}{3} \right)^n\right\rfloor-k,$$ where $k=1$ or $k=2$ for all $n \ge 2$.
\end{proof}

\begin{proposition}\label{proposition:FractionalParts}
For all natural numbers $n \ge 2$, where $k$ is as in Proposition~\ref{proposition:Floor}, we have that $$k-1&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k-1+\varepsilon(n).$$
\end{proposition}
\begin{proof}
From Proposition~\ref{proposition:Floor} we have that $$\left\lfloor \frac{1}{\zeta(n)-1} \right\rfloor=2^n-\left\lfloor \left( \frac{4}{3} \right)^n \right\rfloor-k,$$ and therefore,
$$\left\lfloor \frac{1}{\zeta(n)-1} \right\rfloor-2^n+\left\lfloor \left( \frac{4}{3} \right)^n \right\rfloor+2=2-k.$$
Since the integral part plus the fractional part is the number itself, we have that
\begin{equation}
\label{eq:6}
\frac{1}{\zeta(n)-1}-2^n+ \left( \frac{4}{3} \right)^n +2=\left\{ \frac{1}{\zeta(n)-1} \right\} +\left\{ \left( \frac{4}{3} \right)^n \right\} +2-k.
\end{equation}
We know from Proposition~\ref{proposition:Lower} and Proposition~\ref{proposition:Upper} that $$1&lt; \frac{1}{\zeta(n)-1} -2^n+\left( \frac{4}{3} \right)^n +2&lt;1+\varepsilon(n).$$
Substituting \eqref{eq:6} into this inequality we obtain that $$1&lt;\left\{ \frac{1}{\zeta(n)-1} \right\} +\left\{ \left( \frac{4}{3} \right)^n \right\} +2-k&lt;1+\varepsilon(n).$$
We conclude that $$k-1&lt;\left\{ \frac{1}{\zeta(n)-1} \right\} +\left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k-1+\varepsilon(n).$$
\end{proof}

\begin{proposition}\label{proposition:k=1}
Let $k$ be as in Proposition~\ref{proposition:Floor}. Then $k=1$ occurs finitely many times.
\end{proposition}
\begin{proof}
We know from Proposition~\ref{proposition:FractionalParts} that
$$k-1&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k-1+\varepsilon(n).$$
It follows from Mahler's work \cite{Ma}
that if $p&gt;q \ge 2$ are coprime integers and $\varepsilon&gt;0$, then
$$\left\{ \left(\frac{p}{q} \right)^n \right\}&gt;e^{-\varepsilon n},$$ for all integers $n$, except for at most a finite number of exceptions.

As $\varepsilon(n)=\mathcal{O}\left( \left(\frac{8}{9} \right)^n \right)$, by taking $p=4$, $q=3$ and $\varepsilon=\log(\frac{10}{9})$ we obtain that only a finite number of $n$ satisfy
$$0&lt; \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;\varepsilon(n).$$
Therefore, only a finite number of $n$ satisfy
$$0&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;\varepsilon(n).$$
Thus, $k=1$ occurs a finite number of times.
\end{proof}
The best effective lower bound for the distance of $\left( \frac{4}{3} \right)^n$ to the nearest integer we are aware of is
$$\left\| \left( \frac{4}{3} \right)^n \right\| &gt; 0.4910^n \quad \mbox{for} \quad n \ge 5\,868\,122\,745\,713\,241\,570,$$
given by Pupyrev \cite{Pu}, where $\left\| \cdot \right\|$ denotes the distance to the nearest integer. However this bound is not strong enough for our purposes. Therefore, Proposition~\ref{proposition:k=1} is not effective.

\begin{proof}[Proof of Theorem~\ref{theorem:Main}.]
Theorem~\ref{theorem:Main} follows from combining both Proposition~\ref{proposition:Floor} and Proposition~\ref{proposition:k=1}. 
\end{proof}

\begin{proof}[Proof of Corollary~\ref{corollary:FractionalParts}.]
Corollary~\ref{corollary:FractionalParts} follows from combining both Proposition~\ref{proposition:FractionalParts} and Proposition~\ref{proposition:k=1}.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{corollary:Egypt}.]
We prove this by contradiction. Suppose $\zeta(n)=1 + \frac{1}{m}$ for some integer $m$. Therefore,
$\frac{1}{\zeta(n)-1} =m$ is an integer and thus, $\left\{ \frac{1}{\zeta(n)-1} \right\}=0$. From Proposition~\ref{proposition:FractionalParts} we know that $$k-1&lt;\left\{ \frac{1}{\zeta(n)-1} \right\} +\left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k-1+\varepsilon(n).$$
Therefore, $$k-1&lt;\left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k-1+\varepsilon(n).$$
When $k=2$ we have $$1&lt;\left\{ \left( \frac{4}{3} \right)^n \right\}&lt;1+\varepsilon(n).$$
However, this implies that a fractional part is greater than $1$ which by definition is impossible. This is a contradiction. Therefore, if  $\zeta(n)=1 + \frac{1}{m}$ for some integer $m$, then $k=1$. Thus, this happens finitely many times.
\end{proof}
 
\begin{proposition} \label{proposition:Gen}
For all $\frac{1}{2}&lt;x&lt; \frac{3}{4}$ and $n$ large enough, $$-\left( \frac{4x}{3} \right)^n -x^n-k&lt; \left \{ \frac{x^n}{\zeta(n)-1} \right \} -\left \{ \left(2x \right)^n \right \} &lt; \varepsilon_x(n)-\left( \frac{4x}{3} \right)^n -x^n-k,$$ where $k=0$ or $k=-1$. When $x$ is rational
$k=0$ except for a finite number of exceptions.
\end{proposition}
\begin{proof}
From Proposition~\ref{proposition:Lower} and Proposition~\ref{proposition:Upper} we know that 
$$1&lt; \frac{1}{\zeta(n)-1} -2^n+\left( \frac{4}{3} \right)^n +2&lt; 1+ \varepsilon(n).$$
So
$$0&lt; \frac{1}{\zeta(n)-1} -2^n+\left( \frac{4}{3} \right)^n +1&lt; \varepsilon(n).$$
Recall that $\varepsilon_x(n)=(2x)^n \left( \left( \frac{2}{3} \right)^n + \left( \frac{1}{2} \right)^n \right)^2$. It is obvious that $x^n \varepsilon_y(n)= \varepsilon_{xy}(n)$. Therefore, 
$$0&lt; \frac{x^n}{\zeta(n)-1} -\left(2x \right)^n +\left( \frac{4x}{3} \right)^n + x^n &lt; \varepsilon_x(n),$$ and thus,
\begin{equation}
\label{eq:7}
-\left( \frac{4x}{3} \right)^n-x^n&lt; \frac{x^n}{\zeta(n)-1}-\left(2x \right)^n&lt;\varepsilon_x(n)-\left( \frac{4x}{3} \right)^n-x^n.
\end{equation}
Applying Lemma~\ref{lemma:Floor} twice to \eqref{eq:7} and noticing that $\left\lfloor \left(2x \right)^n \right\rfloor$ appears with a negative sign we obtain that 
$$-1-\left( \frac{4x}{3} \right)^n-x^n&lt;\left\lfloor \frac{x^n}{\zeta(n)-1} \right\rfloor-\left\lfloor \left(2x \right)^n \right\rfloor&lt;1+\varepsilon_x(n)-\left( \frac{4x}{3} \right)^n-x^n.$$ 
Notice that for $\frac{1}{2}&lt;x&lt;\frac{3}{4}$ both bounds of the above inequality tend to $-1$ and $1$ respectively. Also notice that for $n$ large enough we have that 
$$\varepsilon_x(n)-\left( \frac{4x}{3} \right)^n-x^n=x^n \left (\varepsilon(n)-\left( \frac{4}{3} \right)^n-1 \right)&lt;0.$$
As $\left\lfloor \frac{x^n}{\zeta(n)-1} \right\rfloor-\left\lfloor \left(2x \right)^n \right\rfloor$ is an integer, for $n$ large enough we have that 
$$\left\lfloor \frac{x^n}{\zeta(n)-1} \right\rfloor-\left\lfloor \left(2x \right)^n \right\rfloor =-1 \textrm{ or } 0.$$
Therefore, 
$$\frac{x^n}{\zeta(n)-1} -\left(2x \right)^n=k+\left\{ \frac{x^n}{\zeta(n)-1} \right\}-\left\{ \left(2x \right)^n \right\},$$ where $k=-1$ or $k=0$.
Substituting this into \eqref{eq:7} we get that
$$-\left( \frac{4x}{3} \right)^n-x^n&lt; k+\left\{ \frac{x^n}{\zeta(n)-1} \right\}-\left\{ \left(2x \right)^n \right\}&lt;\varepsilon_x(n)-\left( \frac{4x}{3} \right)^n-x^n.$$
Hence, $$-\left( \frac{4x}{3} \right)^n-x^n-k&lt; \left\{ \frac{x^n}{\zeta(n)-1} \right\}-\left\{ \left(2x \right)^n \right\}&lt;\varepsilon_x(n)-\left( \frac{4x}{3} \right)^n-x^n-k.$$
One can use the same argument as in Proposition~\ref{proposition:k=1} to show that for a rational number $\frac{1}{2}&lt;x&lt;\frac{3}{4}$ we have that  $k=-1$ occurs finitely many times.
\end{proof}

One can apply  the same methods for other values of $x$. For $0 &lt; x \le \frac{1}{2}$, both fractional parts converge to $0$, so it is not interesting. For $x= \frac{3}{4}$, one can achieve the same result but with $k=-2$ or $k=-1$. For $\frac{3}{4}&lt;x$, if we want the bounds to converge, there are two possibilities. The first is that there will be more than two fractional parts (apart for some exceptions like $x=1$), and $k$ will take more values, that is, the number of fractional parts. The second is that there will still be two fractional parts and $k$ can only take two values, but one of the fractional parts will be of a finite sum of real numbers to natural powers rather than just one term. For example when $x= \frac{11}{10}$ we have that
$$-k&lt; \left\{ \frac{ \left( \frac{11}{10} \right)^n}{\zeta(n)-1} \right\}-\left\{ \left( \frac{11}{5} \right)^n \right\}+\left\{ \left( \frac{22}{15} \right)^n \right\}+\left\{ \left( \frac{11}{10} \right)^n \right\}&lt;-k+\varepsilon_{\frac{11}{10}}(n),$$ where $k=-2$ or $k=-1$ or $k=0$ or $k=1$. 
Or
$$-k&lt;\left\{ \frac{ \left( \frac{11}{10} \right)^n}{\zeta(n)-1} \right\} - \left\{ \left( \frac{11}{5} \right)^n - \left( \frac{22}{15} \right)^n - \left( \frac{11}{10} \right)^n \right\} &lt;-k+\varepsilon_{\frac{11}{10}}(n),$$ where $k=0$ or $k=1$.

\begin{proof}[Proof of Theorem ~\ref{theorem:Zeta}]
When $x= \frac{2}{3}$ in Proposition~\ref{proposition:Gen} we have
$$-\left( \frac{8}{9} \right)^n-\left( \frac{2}{3} \right)^n-k&lt; \left\{ \frac{ \left( \frac{2}{3} \right)^n}{\zeta(n)-1} \right\}-\left\{ \left( \frac{4}{3} \right)^n \right\}&lt;$$
$$\varepsilon_{\frac{2}{3}}(n)-\left( \frac{8}{9} \right)^n-\left( \frac{2}{3} \right)^n-k,$$
where $k=-1$ or $k=0$.
We know from Corollary~\ref{corollary:FractionalParts} that
$$k'-1&lt; \left\{  \frac{1}{\zeta(n)-1} \right\} + \left\{ \left( \frac{4}{3} \right)^n \right\}&lt;k'-1+\varepsilon(n),$$
where $k'=1$ or $k'=2$.  Adding the two inequalities we get
$$m-\left( \frac{8}{9} \right)^n-\left( \frac{2}{3} \right)^n&lt;\left\{  \frac{1}{\zeta(n)-1} \right\} +\left\{ \frac{ \left( \frac{2}{3} \right)^n}{\zeta(n)-1} \right\}&lt;$$
$$m+\varepsilon(n)+\varepsilon_{\frac{2}{3}}(n)-\left( \frac{8}{9} \right)^n-\left( \frac{2}{3} \right)^n,$$
where $m=0$ or $m=1$ or $m=2$. We can see that $m=0$ or $m=2$ occur finitely many times. Thus, for $n$ large enough we always have $m=1$. This proves the theorem.
\end{proof}


\section{Acknowledgments}
I would like thank James McKee for his alternative proof for
Proposition~\ref{proposition:Upper} using a geometric expansion. This
improvement turned out to be crucial for this paper. I would also like
to thank Jan-Christoph Schlage-Puchta for his helpful comments and
discussions on previous drafts of this paper, as well as writing a
Mathematica program. In addition, I thank Vsevolod Lev for a helpful
suggestion on a previous draft and for Yann Bugeaud for pointing out
Mahler's theorem, which made this paper much stronger. Finally I would
like to thank my father, Yiftach Barnea, for helping me understand the
previous literature, and with writing and organizing this paper.

\begin{thebibliography}{9}

\bibitem{Ma} K.~Mahler, On the fractional parts of the powers of a rational number II, {\em Mathematika} {\bf 4} (1957), 122--124.

\bibitem{Pu} Yu.~A.~Pupyrev, Effectivization of a lower bound for $\|(4/3)^k\|$, {\em Mathematical Notes} {\bf 85} (2009), 877--885.

\bibitem{OEIS} N.~J.~A. Sloane et al., \emph{The On-Line Encyclopedia of Integer
Sequences}, available at 
\url{https://oeis.org}.

\bibitem{Vi} T.~Vijayaraghavan, On the fractional parts of the powers of a number I, {\em J. London Math. Soc.} {\bf 15} (1940), 159--160.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11J71; Secondary 11A55, 11M99, 11Y65, 11K31.

\noindent \emph{Keywords:} continued fraction, fractional part, zeta
function, rational power.

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\noindent (Concerned with sequence
\seqnum{A013697}.)

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\vspace*{+.1in}
\noindent
Received November 10 2018;
revised version received   March 9 2019.
Published in {\it Journal of Integer Sequences}, May 20 2019.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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