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\begin{document}
\vspace*{-40pt}
\centerline{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7
(2007), \#A30}
\vskip 40pt

\begin{center}
\uppercase{\bf A proof of the continued fraction expansion of
}{$e^{2/s}$}
\vskip 20pt 
\textbf{Takao Komatsu\footnote{This research was supported in part by the
Grant-in-Aid for Scientific research (C) (No.
18540006), the Japan Society for the Promotion of
Science.}}\\
{\smallit Graduate School of Science and Technology, 
Hirosaki University, Hirosaki, 036-8561, Japan}\\[0pt]
\texttt{komatsu@cc.hirosaki-u.ac.jp}\\[0pt]
\end{center}

\vskip30pt \centerline{\smallit Received: 3/13/07,
Accepted: 6/6/07, Published: 6/19/07
} % We will fill in the dates
\vskip30pt




\vskip30pt

\centerline{\bf Abstract} 
We give a proof of the continued
 fraction expansion of $e^{2/s}$, where $s\ge 3$ is an odd integer, by
expressing the error between $e^{2/s}$ and its each convergent explicitly
in terms of integrals. 



\pagestyle{myheadings} 
\markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007),
\#A30\hfill} %

\thispagestyle{empty} \baselineskip=15pt \vskip 30pt





\section{Introduction}
Let
$\alpha=[a_0;a_1,a_2,\dots]$ denote the simple continued fraction
expansion of a real $\alpha$, where 
\begin{align*}
\alpha&=a_0+1/\alpha_1,& a_0&=\fl{\alpha},\\
\alpha_n&=a_n+1/\alpha_{n+1},& a_n&=\fl{\alpha_n}\quad(n\ge 1)\,. 
\end{align*} 
The $n$-th convergent of the continued fraction expansion is denoted by $p_n/q_n=[a_0;a_1,\dots,a_n\,]$, and $p_n$ and $q_n$ satisfy the recurrence relation: 
\begin{align*}
p_n&=a_n p_{n-1}+p_{n-2}\quad(n\ge 0),& p_{-1}&=1, &p_{-2}&=0,\\
q_n&=a_n q_{n-1}+q_{n-2}\quad(n\ge 0),& q_{-1}&=0, &q_{-2}&=1\,.
\end{align*}

An irrational number $\alpha$ is well approximated by its $n$-th convergent $p_n/q_n$. In fact, for $n\ge 0$ 
$$
\frac{1}{q_n(q_{n+1}+q_n)}<\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{q_n q_{n+1}}  
$$
(\cite[p. 20]{K}). 
Precisely speaking, by using the algorithm mentioned above, we can 
express (\cite[Lemma 5.4]{B}) the error as 
$$
\alpha-\frac{p_n}{q_n}=\frac{(-1)^n}{q_n(\alpha_{n+1}q_n+q_{n-1})}.
$$

Osler \cite{O} gave a remarkable proof of the simple continued fraction 
$$
e^{1/s}=[1;\overline{(2k-1)s-1,1,1}]_{k=1}^\infty\quad(s\ge 2)
$$
by expressing this error explicitly in terms of integrals. Namely, when $p_n/q_n$ is the $n$-th convergent of the continued fraction of $e^{1/s}$, he showed that for $n\ge 0$ 
\begin{align*}
\frac{p_{3n}}{q_{3n}}-e^{1/s}&=-\frac{1}{q_{3n}}\int_0^1\frac{x^n(x-1)^n}{s^{n+1}n!}e^{x/s} dx\,,\\
\frac{p_{3n+1}}{q_{3n+1}}-e^{1/s}&=\frac{1}{q_{3n+1}}\int_0^1\frac{x^{n+1}(x-1)^n}{s^{n+1}n!}e^{x/s} dx
\end{align*}
and 
$$
\frac{p_{3n+2}}{q_{3n+2}}-e^{1/s}=\frac{1}{q_{3n+2}}\int_0^1\frac{x^n(x-1)^{n+1}}{s^{n+1}n!}e^{x/s} dx\,. 
$$ 
This was the direct extension of the result given by Cohn \cite{C}  concerning $e$. A similar expression can be seen in \cite{D} too. 

It is known that the continued fraction expansion of $e^{2/s}$ is given by 
$$
e^{2/s}=\bigl[1;\overline{\frac{(6k-5)s-1}{2}, (12k-6)s, \frac{(6k-1)s-1}{2}, 1,
1}\,\bigr]_{k=1}^\infty\,,   
$$
\noindent
where $s>1$ is odd (See \cite{P}, \S 32, (2)).  
We shall give another proof of the continued fraction expansion of $e^{2/s}$ by showing similar errors explicitly. 
For convenience, for $n\ge 0$ put 
\begin{align*}
A_n&=\left(\frac{2}{s}\right)^{3n+1}\int_0^1\frac{x^{3n}(x-1)^{3n}}{(3n)!}e^{2x/s}dx\,,\\
B_n&=\frac{2^{3n+1}}{s^{3n+2}}\int_0^1\frac{x^{3n+1}(x-1)^{3n+1}}{(3n+1)!}e^{2x/s}dx\,,\\
C_n&=\left(\frac{2}{s}\right)^{3n+3}\int_0^1\frac{x^{3n+2}(x-1)^{3n+2}}{(3n+2)!}e^{2x/s}dx\,,\\
D_n&=\left(\frac{2}{s}\right)^{3n+3}\int_0^1\frac{x^{3n+3}(x-1)^{3n+2}}{(3n+2)!}e^{2x/s}dx\,,
\end{align*}
and 
$$
E_n=\left(\frac{2}{s}\right)^{3n+3}\int_0^1\frac{x^{3n+2}(x-1)^{3n+3}}{(3n+2)!}e^{2x/s}dx\,.
$$
Then, our main theorem is stated as follows. 

\newpage
\begin{theorem} 
Let $p_n/q_n$ be the $n$-th convergent of the continued fraction of $e^{2/s}$. Then, for $n\ge 0$   
\begin{align*}
\frac{p_{5n}}{q_{5n}}-e^{2/s}&=-\frac{1}{q_{5n}}A_n\,,\\
\frac{p_{5n+1}}{q_{5n+1}}-e^{2/s}&=-\frac{1}{q_{5n+1}}B_n\,,\\
\frac{p_{5n+2}}{q_{5n+2}}-e^{2/s}&=-\frac{1}{q_{5n+2}}C_n\,,\\
\frac{p_{5n+3}}{q_{5n+3}}-e^{2/s}&=\frac{1}{q_{5n+3}}D_n\,,
\end{align*}
and 
$$
\frac{p_{5n+4}}{q_{5n+4}}-e^{2/s}=\frac{1}{q_{5n+4}}E_n\,.
$$
\end{theorem}


\section{The Continued Fraction of $e^{2/s}$} 


The proof of the main theorem is based upon the following identities. 

\begin{lemma} 
\begin{align}
A_n&=-E_{n-1}-D_{n-1}\,,\label{1}\\
B_n&=\frac{(6n+1)s-1}{2}A_n-E_{n-1}\,,\label{2}\\
C_n&=(12n+6)s B_n+A_n\,,\label{3}\\
D_n&=-\frac{(6n+5)s-1}{2}C_n-B_n\,,\label{4}
\end{align}
and 
\begin{equation}
E_n=D_n-C_n\,.\label{5}
\end{equation} 
\end{lemma}

These identities correspond with the desired relations: 
\begin{align*}
p_{5n}&=p_{5n-1}+p_{5n-2},&\quad q_{5n}&=q_{5n-1}+q_{5n-2},\\
p_{5n+1}&=\frac{(6n+1)s-1}{2}p_{5n}+p_{5n-1},&\quad q_{5n+1}&=\frac{(6n+1)s-1}{2}q_{5n}+q_{5n-1},\\
p_{5n+2}&=(12n+6)s p_{5n+1}+p_{5n},&\quad q_{5n+2}&=(12n+6)s q_{5n+1}+q_{5n},\\
p_{5n+3}&=\frac{(6n+5)s-1}{2}p_{5n+2}+p_{5n+1},&\quad q_{5n+3}&=\frac{(6n+5)s-1}{2}q_{5n+2}+q_{5n+1},\\
p_{5n+4}&=p_{5n+3}+p_{5n+2},&\quad q_{5n+4}&=q_{5n+3}+q_{5n+2}. 
\end{align*}

We shall prove Theorem 1.1 and Lemma 2.1 simultaneously. 


\noindent
{\it Proof.}
When $n=0$, the relations in Theorem 1.1 are true. Indeed, 
$$
A_0=\frac{2}{s}\int_0^1 e^{2x/s}dx=e^{2/s}-1=q_0 e^{2/s}-p_0\,.
$$
\begin{align*}
B_0&=\frac{2}{s^2}\int_0^1 x(x-1)e^{2x/s}dx=\frac{1}{2s}\bigl[\bigl(2x^2-2(s+1)x+s^2+s\bigr)e^{2x/s}\bigr]_0^1\\
&=\frac{s-1}{2}e^{2/s}-\frac{s+1}{2}=q_1 e^{2/s}-p_1\,. 
\end{align*}  
\begin{align*}
C_0&=\left(\frac{2}{s}\right)^3\frac{1}{2}\int_0^1 x^2(x-1)^2 e^{2x/s}dx\\
&=\frac{1}{s^2}\bigl[\bigl(2x^4-4(s+1)x^3+2(3s^2+3s+1)x^2\\
&\qquad -2s(3s^2+3s+1)x+3s^4+3s^3+s^2\bigr)e^{2x/s}\bigr]_0^1\\
&=(3s^2-3s+1)e^{2/s}-(3s^2+3s+1)=q_2 e^{2/s}-p_2\,. 
\end{align*}
\begin{align*}
D_0&=\left(\frac{2}{s}\right)^3\frac{1}{2}\int_0^1 x^3(x-1)^2 e^{2x/s}dx\\
&=-\frac{1}{2s^2}\bigl[\bigl(4x^5-(10s+8)x^4+(20s^2+16s+4)x^3-(30s^3+24s^2+6s)x^2\\
&\qquad +(30s^4+24s^3+6s^2)x-(15s^5+12s^4+3s^3)\bigr)e^{2x/s}\bigr]_0^1\\
&=\frac{15s^3}{2}+6s^2+\frac{3s}{2}-\left(\frac{15s^3}{2}-9s^2+\frac{9s}{2}-1\right)e^{2/s}=p_3-q_3 e^{2/s}\,. 
\end{align*}
\begin{align*}
E_0&=\left(\frac{2}{s}\right)^3\frac{1}{2}\int_0^1 x^2(x-1)^3 e^{2x/s}dx\\
&=-\frac{1}{2s^2}\bigl[\bigl(4x^5-(10s+12)x^4+(20s^2+24s+12)x^3-(30s^3+36s^2+18s+4)x^2\\
&\qquad +(30s^4+36s^3+18s^2+4s)x-(15s^5+18s^4+9s^3+2s^2)\bigr)e^{2x/s}\bigr]_0^1\\
&=\frac{15s^3}{2}+9s^2+\frac{9s}{2}+1-\bigl(\frac{15s^3}{2}-6s^2+\frac{3s}{2}\bigr)e^{2/s}=p_4-q_4 e^{2/s}\,. 
\end{align*}
 
Suppose that Theorem 1.1 is true up to some integer $n-1(\ge 0)$. 
Since 
$$
\frac{d}{dx}\bigl(x^{3n}(x-1)^{3n}e^{2x/s}\bigr)=3n x^{3n-1}(x-1)^{3n}e^{2x/s}+3n x^{3n}(x-1)^{3n-1}+\frac{2}{s}x^{3n}(x-1)^{3n}e^{2x/s}\,, 
$$ 
by integrating from $0$ to $1$ we get (\ref{1}). Hence, 
\begin{align*}
p_{5n}-q_{5n}e^{2/s}&=(p_{5n-1}+p_{5n-2})-(q_{5n-1}+q_{5n-2})e^{2/s}\\
&=E_{n-1}+D_{n-1}=-A_n\,. 
\end{align*}

Since 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=3n x^{3n-1}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n}(x-1)^{3n+1}e^{2x/s}\\
&=-3n x^{3n-1}(x-1)^{3n}e^{2x/s}+\bigl(6n+1-\frac{2}{s}\bigr)x^{3n}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+1}(x-1)^{3n}e^{2x/s} 
\end{align*} 
and 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+1}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=(3n+1)x^{3n}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+1}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+1}(x-1)^{3n+1}e^{2x/s}\\
&=2(3n+1)x^{3n+1}(x-1)^{3n}e^{2x/s}-(3n+1)x^{3n}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+1}(x-1)^{3n+1}e^{2x/s}\,,  
\end{align*} 
by integrating from $0$ to $1$ and canceling the term of $x^{3n+1}(x-1)^{3n}e^{2x/s}$ we get 
\begin{multline*}
\frac{s}{2}(-3n)\int_0^1 x^{3n-1}(x-1)^{3n}e^{2x/s}dx+\frac{(6n+1)s-1}{2}\int_0^1 x^{3n}(x-1)^{3n}e^{2x/s}dx\\
-\frac{1}{s(3n+1)}\int_0^1 x^{3n+1}(x-1)^{3n+1}e^{2x/s}dx=0\,. 
\end{multline*} 
Thus, we have (\ref{2}). Hence, 
\begin{align*}
p_{5n+1}-q_{5n+1}e^{2/s}&=\left(\frac{(6n+1)s-1}{2}p_{5n}+p_{5n-1}\right)-\left(\frac{(6n+1)s-1}{2}q_{5n}+q_{5n-1}\right)e^{2/s}\\
&=-\frac{(6n+1)s-1}{2}A_n+E_{n-1}=-B_n\,. 
\end{align*} 

Notice that 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+2}(x-1)^{3n+2}e^{2x/s}\bigr)\\
&=(3n+2)x^{3n+1}(x-1)^{3n+2}e^{2x/s}+(3n+2)x^{3n+2}(x-1)^{3n+1}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+2}e^{2x/s}\\
&=(6n+4)x^{3n+2}(x-1)^{3n+1}e^{2x/s}-(3n+2)x^{3n+1}(x-1)^{3n+1}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+2}e^{2x/s}\,,  
\end{align*} 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+2}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=(3n+2)x^{3n+1}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+2}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+1}e^{2x/s}\\
&=(6n+3)x^{3n+1}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+1}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+1}e^{2x/s} 
\end{align*} 
and 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+1}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=2(3n+1)x^{3n+1}(x-1)^{3n}e^{2x/s}-(3n+1)x^{3n}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+1}(x-1)^{3n+1}e^{2x/s}\,.  
\end{align*} 

Integrating these three equations from $0$ to $1$ and canceling the terms of $x^{3n+2}(x-1)^{3n+1}e^{2x/s}$ and $x^{3n+1}(x-1)^{3n}e^{2x/s}$, we get 
\begin{multline*}
\left(\frac{2}{s}\right)^2\int_0^1 x^{3n+2}(x-1)^{3n+2}e^{2x/s}dx\\
=(12n+6)(3n+2)\int_0^1 x^{3n+1}(x-1)^{3n+1}e^{2x/s}dx+(3n+2)(3n+1)\int_0^1 x^{3n}(x-1)^{3n}e^{2x/s}dx\,. 
\end{multline*} 
Thus, we have (\ref{3}). Hence, 
\begin{align*}
p_{5n+2}-q_{5n+2}e^{2/s}&=\bigl((12n+6)s p_{5n+1}+p_{5n}\bigr)-\bigl((12n+6)s q_{5n+1}+q_{5n}\bigr)e^{2/s}\\
&=-(12n+6)s B_n-A_n=-C_n\,. 
\end{align*} 


Notice that 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+3}(x-1)^{3n+2}e^{2x/s}\bigr)\\
&=(3n+3)x^{3n+2}(x-1)^{3n+2}e^{2x/s}+(3n+2)x^{3n+3}(x-1)^{3n+1}e^{2x/s}+\frac{2}{s}x^{3n+3}(x-1)^{3n+2}e^{2x/s}\,,  
\end{align*} 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+3}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=(3n+3)x^{3n+2}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+3}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+3}(x-1)^{3n+1}e^{2x/s}\\
&=\left(6n+4+\frac{2}{s}\right)x^{3n+2}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+2}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+2}e^{2x/s}\,,  
\end{align*} 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+2}(x-1)^{3n+2}e^{2x/s}\bigr)\\
&=\left(6n+4-\frac{2}{s}\right)x^{3n+2}(x-1)^{3n+1}e^{2x/s}-(3n+2)x^{3n+1}(x-1)^{3n+1}e^{2x/s}+\frac{2}{s}x^{3n+3}(x-1)^{3n+1}e^{2x/s} 
\end{align*} 
and 
\begin{align*}
&\frac{d}{dx}\bigl(x^{3n+2}(x-1)^{3n+1}e^{2x/s}\bigr)\\
&=(3n+2)x^{3n+1}(x-1)^{3n+1}e^{2x/s}+(3n+1)x^{3n+2}(x-1)^{3n}e^{2x/s}+\frac{2}{s}x^{3n+2}(x-1)^{3n+1}e^{2x/s}\,.  
\end{align*} 

Integrating these four equations from $0$ to $1$ and eliminating the terms of $x^{3n+3}(x-1)^{3n+1}e^{2x/s}$, $x^{3n+2}(x-1)^{3n+1}e^{2x/s}$ and $x^{3n+2}(x-1)^{3n}e^{2x/s}$, we get 
\begin{multline*}
\frac{4}{s}\int_0^1 x^{3n+3}(x-1)^{3n+2}e^{2x/s}dx\\
=-\left(12n+10-\frac{2}{s}\right)\int_0^1 x^{3n+2}(x-1)^{3n+2}e^{2x/s}dx-(3n+2)\int_0^1 x^{3n+1}(x-1)^{3n+1}e^{2x/s}dx\,. 
\end{multline*} 
Thus, we have (\ref{4}). Hence, 
\begin{align*}
p_{5n+3}-q_{5n+3}e^{2/s}&=\left(\frac{(6n+5)s-1}{2}p_{5n+2}+p_{5n+1}\right)-\left(\frac{(6n+5)s-1}{2}q_{5n+2}+q_{5n+1}\right)e^{2/s}\\
&=-\frac{(6n+5)s-1}{2}C_n-B_n=D_n\,. 
\end{align*} 


Since $x^{3n+3}(x-1)^{3n+2}-x^{3n+2}(x-1)^{3n+2}=x^{3n+2}(x-1)^{3n+3}$, we get $D_n-C_n=E_n$, which is (\ref{5}). Hence, 
\begin{align*}
p_{5n+4}-q_{5n+4}e^{2/s}=(p_{5n+3}+p_{5n+2})-(q_{5n+3}+q_{5n+2})e^{2/s}
=D_n-C_n=E_n\,. 
\end{align*}
\hfill $\Box$


\section{The Continued Fraction of $e^2$}

Let $\frac{p_n^\ast}{q_n^\ast}$ be the $n^{\mathrm{th}}$ convergent of the
continued fraction of
$e^2=[7;\overline{3k-1,1,1,3k,12k+6}~]_{k=1}^\infty$. Then, for $n\ge 0$
we have 
$$
\frac{p_n^\ast}{q_n^\ast}=\frac{p_{n+2}}{q_{n+2}}\,, 
$$
where $p_n/q_n$ is the $n$-th convergent of the continued fraction of $e^{2/s}$ mentioned above. Thus, by replacing $p_n/q_n$ by $p_{n-2}^\ast/q_{n-2}^\ast$ ($n\ge 2$), Theorem 1.1 with Lemma 2.1 holds for $s=1$. 


\section{Additional Comments} 

Some results in our theorem can be derived directly from Osler's results. By the relation for $i=1,2,\dots$ 
$$
[\,\dots,1,a_i-\frac{1}{2},1,\dots\,]=[\,\dots,1,a_{3i-2},4a_{3i-1}+2,a_{3i},1,\dots\,]\,, 
$$
if we replace $s$ by $s/2$ in the continued fraction $[1;\overline{(2k-1)s-1,1,1}]_{k=1}^\infty$, then we have 
\begin{align*}
[1;\overline{(2k-1)s/2-1,1,1}]_{k=1}^\infty
&=\bigl[1;\overline{ks-\frac{s+1}{2}-\frac{1}{2},1,1}\bigr]_{k=1}^\infty\\ 
&=\bigl[1;\overline{\frac{(6k-5)s-1}{2}, (12k-6)s, \frac{(6k-1)s-1}{2}, 1, 1}\,\bigr]_{k=1}^\infty\,. 
\end{align*}
Therefore, if we replace $s$ by $s/2$ and $n$ by $3n$ in the Osler's integral for $p_{3n}-q_{3n}e^{1/s}$, we get our integral for $p_{5n}-q_{5n}e^{2/s}$. If we replace $s$ by $s/2$ and $n$ by $3n+2$ in the Osler's integral for $p_{3n+1}-q_{3n+1}e^{1/s}$, we get our integral for $p_{5n+3}-q_{5n+3}e^{2/s}$.  If we replace $s$ by $s/2$ and $n$ by $3n+2$ in the Osler's integral for $p_{3n+2}-q_{3n+2}e^{1/s}$, we get our integral for $p_{5n+4}-q_{5n+4}e^{2/s}$.  


\section{Acknowledgement} 

This work was initiated when the author stayed in the Department of Mathematics and Statistics, University of Tennessee at Martin in 2006. In special, he thanks Sarah Holliday, Bill Austin, Louis Kolitsch, and Chris Caldwell. The author also thanks the anonymous referee for some helpful comments. 



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\end{document}

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