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\begin{center}
\vskip 1cm{\LARGE\bf Infinite Products Involving $\zeta(3)$ and \\
\vskip .1in
Catalan's  Constant}
\vskip 1cm 

\large
Yasuyuki Kachi  \\
Department of Mathematics \\
University of Kansas  \\
Lawrence, KS 66045-7523 \\
USA   \\
\href{mailto:kachi@math.ku.edu}{\tt kachi@math.ku.edu} \\
\ \\
Pavlos Tzermias \\
Department of Mathematics  \\
University of Patras  \\
26500 Rion (Patras) \\
Greece  \\
\href{mailto:tzermias@math.upatras.gr}{\tt tzermias@math.upatras.gr} \\
\end{center}


\newcommand{\Q}{{\mathbb Q}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\F}{{\mathbb F}}

\vskip .2in

\begin{abstract} 
We present some  
infinite product formulas for 
$e^{\frac{7 \zeta (3)}{\pi^2}}$, $e^{\frac{4G}{\pi}}$  
and 
$e^{\frac{2 G}{\pi} \ \pm \frac{1}{2}}$,  
where $G$ is Catalan's constant. We relate these formulas to similar 
ones obtained by   
Guillera and Sondow in the context of their systematic study of 
Lerch's transcendent. Our proofs are entirely elementary.  
\end{abstract} 


\section{Introduction}  

This paper studies some infinite product formulas involving two classical  
constants, namely 
$\zeta \big(3\big)$ and  
Catalan's constant, whose definition we now recall:
$$
\zeta \big(3\big)   
\ = \  
\sum\limits_{n = 1}^{\infty} 
\frac{1}{n^3} \   
$$
and  
$$
G 
\ = \  
\sum\limits_{m = 0}^{\infty} 
\ \frac{\big(-1\big)^m}  
{\big(2 m +  1\big)^2} 
 \ . $$
The following formulas are reminiscent of similar formulas 
obtained by Guillera and Sondow in \cite{GS}:  

\begin{proposition} \label{zeta3} 
The following formulas hold:  
\begin{equation} \label{zeta3i} 
e^{\frac{7 \zeta(3)}{4  \pi^2} + \frac{1}{4} 
}   = \lim_{m \to \infty}   
\prod\limits_{n = 1}^{2m+1} 
\ 
\frac{1}{\sqrt[4]{e}} \ \Big(1 - \frac{1}{n+1}\Big)^{\frac{n(n+1)}{2} (-1)^n}.  
\end{equation}     

\begin{equation} \label{zeta3ii}  
e^{\frac{7 \zeta(3)}{4 \pi^2} -\frac{1}{4}} = \lim_{m \to \infty} 
\prod\limits_{n=1}^{2m} \sqrt[4]{e} \ \Big(1 - \frac{1}{n+1} \Big)^{\frac{n(n+1)}{2} (-1)^n} .   
\end{equation}

\begin{equation} \label{zeta3iii} 
e^{\frac{ 7 \zeta(3)}{\pi^2}} = \lim_{m \to \infty} \Big( 
\frac{ 2^{2^2} \cdot 4^{4^2} \cdot 6^{6^2} \cdots (2m)^{(2m)^2}}
{1^{1^2} \cdot 3^{3^2} \cdot 5^{5^2} \cdots (2m-1)^{(2m-1)^2}}\Big)^4 \ \Big(\frac{(2m+2)^{4m+5}}{(2m+1)^{12m +9}} \Big)^{m} .  
\end{equation} 
\end{proposition} 


\begin{proposition}\label{catalan}  
The following formulas hold: 
\begin{equation} \label{catalani}
e^{\frac{2 G}{\pi} -  
\frac{1}{2} }  =  \lim_{m \to \infty}  
\prod\limits_{n = 1}^{2m} 
\Big( 1 - \frac{2}{2n+1} \Big)^{n (-1)^n} .  
\end{equation}   

\begin{equation} \label{catalanii}
e^{\frac{2G}{\pi}+ \frac{1}{2}} = \lim_{m \to \infty} \prod\limits_{n=1}^{2m+1} \Big( 1 - \frac{2}{2n+1}  \Big)^{n (-1)^n} . 
\end{equation} 

\begin{equation} \label{catalaniii} 
e^{\frac{4G}{\pi}} = \lim_{m \to \infty} \Big( 
\frac{3^3 \cdot 7^7 \cdot 11^{11} \cdots (4m-1)^{4m-1}}{1^1 \cdot 5^5 \cdot 9^9 \cdots (4m-3)^{4m-3}}  \Big)^2 \ \frac{(4m+3)^{2m+1}}{(4m+1)^{6m+1}} .  
\end{equation}
\end{proposition} 
We claim no novelty for the formulas themselves; our only purpose 
here is to present completely 
elementary proofs of these formulas and to establish the not-so-obvious 
facts below: 

\begin{fact}\label{zeta3gs}    
Formula $\eqref{zeta3iii}$ is    
equivalent to the following formula 
given by Guillera and Sondow \cite[Example~5.3]{GS}: 
\begin{equation*}
\begin{aligned} 
e^{\frac{7 \zeta(3)}{4 \pi^2} 
} &amp;=   
e^{  \  
\sum\limits_{n=1}^{\infty} \frac{ 
n (n + 1)}{2^{n+3}} 
\ 
\sum\limits_{k=0}^{n} 
\ (-1)^{k+1} \  
{n \choose k} \ \operatorname{log} (k  + 1)} \\  
&amp;=  \prod_{n=1}^{\infty} \Big(  
\prod_{k=0}^{n} 
\ \Big(k + 1 \Big)^{ (-1)^{k+1} \ {n \choose k}}  
\Big)^{\frac{n (n+1)}{2^{n+3}}}\\  
&amp;=   
\bigg(\frac{2^1}{1^1} 
\bigg)^{\frac{{1 \cdot 2}}{ 2^4}} 
\  
\bigg(\frac{2^2}{1^1 
 \cdot  3^1} \bigg)^{\frac{2 \cdot 3}{ 2^5}} 
\  
\bigg(\frac{2^3  \cdot  4^1 
}{1^1 
 \cdot  3^3} \bigg)^{\frac{3 \cdot 4}{ 2^6}} 
\  
\bigg(\frac{2^4  \cdot 4^4 
}{1^1 
 \cdot  3^6  \cdot \ 5^1}  
\bigg)^{\frac{ 4 \cdot 5 }{ 2^7}} 
 \cdots .  
\end{aligned}
\end{equation*} 
\end{fact} 


\begin{fact}\label{catalangs} 
Formula $\eqref{catalani}$ follows from rearranging the factors of the 
following formula 
given by Guillera and Sondow \cite[Example~5.5]{GS}: 
\begin{equation*}
\begin{aligned} 
e^{\frac{G}{\pi}} &amp;=   
e^{ \    
\sum\limits_{n=1}^{\infty} \frac{n}{2^{n+2}} 
\ 
\sum\limits_{k=0}^{n} 
\ (-1)^{k+1} \  
{n \choose k} \ \operatorname{log} (2  
k  + 1)} \\  
&amp;= \prod_{n=1}^{\infty} \Big(  
\prod_{k=0}^{n} 
\ \Big(2 k  + 1 \Big)^{(-1)^{k+1} \ {n \choose k}} \Big)^{\frac{n}{2^{n+2}}}  
\\ 
&amp;=  
\bigg(\frac{3^1}{1^1} \ 
\bigg)^{\frac{1}{2^3}} 
\  
\bigg(\frac{3^2}{1^1 
 \cdot  5^1} \bigg)^{\frac{2}{ 2^4}} 
\  
\bigg( \frac{3^3  \cdot  7^1 
}{1^1 
 \cdot  5^3} \bigg)^{\frac{3}{2^5}} 
\  
\bigg(\frac{3^4  \cdot 7^4 
}{1^1 
 \cdot  5^6  \cdot  9^1}   
\bigg)^{\frac{4}{2^6}} 
 \cdots ,    
\end{aligned}
\end{equation*}
which in turn is equivalent to formula $\eqref{catalaniii}$. 
\end{fact} 

\section{Proof of Proposition \ref{catalan}} 

We begin with the following formula which is a classically known 
Fourier expansion (see, for
example, Exercise 11.15(c) in \cite[p. \ 338]{A}):

\begin{formula} \label{f21}  
Let $\sigma \in  
\big[-\frac{1}{2}, 
\frac{1}{2} \big] 
\setminus \{0\}$. 
Then 
$$  
\sum\limits_{m=0}^{\infty}
\frac{\operatorname{cos} 
\Big(\pi \big(2  m  + 1\big) \ \sigma 
\Big)}{2 m  + 1} 
\ = \ 
\frac{1}{2} 
\ \operatorname{log}  \bigg| 
\ \operatorname{cot} \ 
\Big(\frac{\pi}{2} \ \sigma \Big) 
\ \bigg|. $$  
\end{formula} 
The following formula, which follows directly from Formula $\ref{f21}$ 
by integrating 
both sides over the interval  
$\Big[0, \ \frac{1}{2} \Big]$,  
is also well-known (see, for example, \cite[p. \ 239]{BB}):   

\begin{formula}\label{f22}  
$$
G \ = \ 
\int_{\theta = 0}^{\pi/4} 
\  
\operatorname{log}  
\Big(\operatorname{cot} \theta \Big) 
\ d  \theta . 
$$
\end{formula} 
By applying integration by parts  
to the latter integral, we obtain 

\begin{corollary}\label{c23}  
$$
G \ = \ 
\frac{1}{2} \  
\int_{\alpha = 0}^{1/2} 
\  
\frac{\pi^2 \ \alpha}{ 
\operatorname{sin} \big(\pi \ \alpha \big) 
} 
\ d  \alpha .
$$
\end{corollary}
The following formula is also well-known (see, for example, 
\cite[p. \ 155]{N}): 

\begin{formula}\label{f24}  
Let $\alpha \in {\mathbb R} \setminus {\mathbb Z}$ 
and $s \in \Big(- \frac{1}{2}, 
\frac{1}{2} \Big)$. 
Then
$$\operatorname{cos} 
\big(2 \pi \alpha s\big) 
\ = \  
\frac{ 
\operatorname{sin} \big(\pi \alpha\big) 
}{\pi} 
\ 
\Big( 
\frac{1}{\alpha} 
\ + \  
2 \alpha \  
\sum_{m = 1}^{\infty} 
\ \frac{\big(-1\big)^m 
}{\alpha^2 
- m^2 } 
\ \operatorname{cos} 
\big(2 \pi m s\big) 
\Big) .$$  
\end{formula} 
Setting $s = 0$ in Formula $\ref{f24}$ gives: 

\begin{corollary}\label{c25}  
Let $\alpha \in {\mathbb R} \setminus {\mathbb Z}$. 
Then 

$$
1 \ = \ 
\frac{ 
\operatorname{sin} \big(\pi \alpha\big) 
}{\pi} 
\ 
\Big( 
\frac{1}{\alpha} 
\ + \ 
2 \alpha \  
\sum_{m = 1}^{\infty} 
\ \frac{\big(-1\big)^m 
}{\alpha^2 
 - m^2 } \Big) .$$  
\end{corollary} 

\begin{lemma}\label{l26}  
Let $m \in {\mathbb Z}$, \ 
$m \geq 1$. Then 
$$
\int_{\alpha = 0}^{1/2} 
\  \frac{\alpha^2}{ 
\alpha^2 - m^2 } 
\ d  \alpha 
\ = \ 
\frac{1}{2} 
\  + \  
\frac{m}{2} 
\ \operatorname{log} 
\ \frac{2 m - 1 
}{2 m  +  1}.
$$
\end{lemma} 
\begin{proof}  
This is straightforward: 

\begin{equation*}
\begin{aligned} 
\int_{\alpha = 0}^{1/2} 
\ \frac{\alpha^2}{ 
\alpha^2  -  m^2 } 
\  d  \alpha 
&amp;=  
\frac{1}{2} \  
\int_{\alpha = 0}^{1/2} 
\  
\bigg( 2 \ + \  
\frac{m}{\alpha  -  m 
} 
\  - \  
\frac{m}{\alpha  +  m 
} \bigg) \ d  \alpha \\   
&amp;=  
\frac{1}{2} 
\  
\Bigg[  
2 \alpha 
\ + \ m 
\ \operatorname{log}  
\Big(- \alpha +  m \Big) 
\  - \  m 
\ \operatorname{log}  
\Big(\alpha  +  m \Big) 
\ \Bigg]_{\alpha = 0}^{1/2} \\  
&amp;=  
\dfrac{1}{2} 
\ + \  
\frac{m}{2} 
\, \, \operatorname{log} 
\ \frac{2 m  -  1 
}{2 m  +  1} .
\end{aligned}
\end{equation*}  
\end{proof}

We now proceed with the proof of   
formula $\eqref{catalani}$.    
By Corollary $\ref{c25}$, we have  
$$
\frac{\pi^2 \alpha}{ 
\operatorname{sin} \big(\pi \alpha\big) 
} \  = \  
\pi 
\  + \  
2 \pi \alpha^2 \  
\sum_{m = 1}^{\infty} 
\ \frac{\big(-1\big)^m 
}{\alpha^2 
 - m^2 } . 
$$
Integrating both sides with respect to $\alpha$ 
over the interval $\Big[0,  
\frac{1}{2} \Big]$ gives  

\begin{equation}\label{e21}  
\int_{\alpha = 0}^{1/2} 
\ 
\frac{\pi^2 \alpha}{ 
\operatorname{sin} \big(\pi \alpha\big) 
} \  d  \alpha 
\ = \ 
\frac{\pi}{2} 
\ + \ 
2 \pi \ 
\int_{\alpha = 0}^{1/2} 
\Big(\sum_{m = 1}^{\infty} 
\ \frac{(-1)^m \  \alpha^2 
}{\alpha^2 
 -  m^2} \Big) \ d  \alpha . 
\end{equation} 
Consider the sequence of functions 
$$ f_m \big(\alpha\big) \ = \  
\frac{\big(-1\big)^m \ \alpha^2 
}{\alpha^2  -  m^2} $$
on the interval  
$I = [0 , \frac{1}{2}]$, 
where $m = 1, 2, \ldots .$ 
Since $\alpha \in \, I$, we clearly have 
$$ \big| f_m \big(\alpha\big) \big| 
\ = \  
\frac{\alpha^2 
}{\big|\alpha^2  -  m^2 \big|}
\ \leq \  
\frac{\frac{1}{4}}{ 
m^2 -  \frac{1}{4} } 
\ = \  
\dfrac{1}{4 m^2 -  1 
} \ \leq \ \dfrac{1}{ 
2 m^2 },
$$  
for all $m$. Since 
$$
\sum_{m=1}^{\infty} \ \frac{1}{2 m^2} 
$$ 
converges, it follows from the 
Weierstrass $M$-test that the series  
$$ 
\sum_{m=1}^{\infty} \ f_m \big(\alpha\big) 
$$ 
converges uniformly on $I$,  and, by  
well-known principles,  (see, 
for example, \cite[Thm.~9.9, p.\ 226]{A}), can 
therefore be integrated term by term. In other words, 
if we set  
$$
a_m \  = \  \big(-1\big)^m \ 
\Big(1 \ 
+ \  m \ \operatorname{log} \ \frac{2 m -  1}{2 m  +  1} \Big) , 
$$
then $\eqref{e21}$ and Lemma $\ref{l26}$ imply that   

\begin{equation}\label{e22}  
\int_{\alpha=0}^{1/2} \ \frac{\pi^2 \alpha}{ 
\operatorname{sin} \big(\pi  \alpha \big)} \  
d \alpha 
\ = \  
\frac{\pi}{2} \ + \  
\pi \ \sum_{m = 1}^{\infty} 
\ a_m. 
\end{equation} 
The left-hand side of $\eqref{e22}$ is a definite integral 
of the  
continuous function 
$\frac{\pi^2 \alpha}{ 
\operatorname{sin} \big(\pi \alpha \big)}$ 
over the interval $[0, \frac{1}{2}]$. Hence 
the left-hand side of $\eqref{e22}$ is a 
real number which implies that  
$$\lim_{m \to \infty} a_m \ = \ 0 . $$ 
Keeping this in mind, define
$$
A_n \ = \ \sum_{m=1}^n \ a_m . 
$$  
Then, by $\eqref{e22}$, we have 

\begin{equation*}
\begin{aligned} 
\int_{\alpha=0}^{1/2}
\ 
\frac{\pi^2  \alpha}{ 
\operatorname{sin} \big( \pi \alpha\big) 
} \ d \alpha 
&amp;=   
\frac{\pi}{2} 
 +  \pi \ \lim_{n \to \infty} A_n  =   
\frac{\pi}{2} 
 +  \pi \ \lim_{N \to \infty} A_{2N}   
\ = \  
\frac{\pi}{2} 
 +  
\pi \  
\lim_{N \to \infty} \sum_{m=1}^{N} (a_{2m-1} +a_{2m}) 
\\   
&amp;=   
\pi \ 
\Big(  
\frac{1}{2} 
\  + \ \lim_{N \to \infty}  
\ \sum_{m=1}^N \ 
\Big( - (2 m -1)  \operatorname{log} 
\frac{4 m  -  3}{ 
4 m -  1} 
\ + \ 2  m \ \operatorname{log}  
\frac{4  m  -  1}{ 
4 m  + 1} \Big) \Big) 
\\ 
&amp;= 
\pi \  
\Big( 
\frac{1}{2} 
\ + \ \lim_{N \to \infty}  
\operatorname{log} \  
\prod_{m = 1}^N 
\ 
\frac{ 
\big(4 m  - 1\big)^{4m-1} 
}{ 
\big(4 m  -  3\big)^{2m-1} \ 
\big(4 m  +  1\big)^{2m}} 
\ \Big) 
\\
&amp;=  
\pi \  
\Big( 
\frac{1}{2} 
\ + \ 
\operatorname{log} \  
\prod_{m = 1}^{\infty} 
\ 
\frac{ 
\big(4 m  - 1\big)^{4m-1} 
}{ 
\big(4 m  - 3\big)^{2m-1} \ 
\big(4 m  +  1\big)^{2m}} 
\ \Big) .
\end{aligned} 
\end{equation*} 
By Corollary $\ref{c23}$, the left-hand side 
equals $2 G$, therefore  
$$
\frac{2 G}{\pi} 
\ - \ \frac{1}{2} 
\ = \ 
\operatorname{log} \  
\prod_{m = 1}^{\infty} 
\  
\frac{ 
\big(4  m  -  1\big)^{4m-1} 
}{ 
\big(4 m  -  3\big)^{2m-1} \ 
\big(4 m  +  1\big)^{2m}} .  
$$
Therefore,

\begin{equation*}
\begin{aligned} 
e^{\frac{2G}{\pi} - \frac{1}{2}} &amp;= \lim_{m\to \infty} 
\frac{3^3}{1^1 \cdot 5^2} \cdot \frac{7^7}{5^3 \cdot 9^4} \cdot \frac{11^{11}}{9^5 \cdot 13^6} \cdots \frac{(4m-1)^{4m-1}}{(4m-3)^{2m-1} (4m+1)^{2m}} \\ 
&amp;= \lim_{m \to \infty} \frac{3^3 \cdot 7^7 \cdot 11^{11} \cdots (4m-1)^{4m-1}}{5^5 \cdot 
9^9 \cdot 13^{13} \cdots (4m-3)^{4m-3} \cdot (4m+1)^{2m}} \\ 
&amp;= \lim_{m \to \infty} \Big(\frac{1}{3}\Big)^{-1} \Big(\frac{3}{5}\Big)^2 
\Big(\frac{5}{7}\Big)^{-3} \cdots \Big(\frac{4m-1}{4m+1}\Big)^{2m} = \lim_{m \to \infty} \prod_{n=1}^{2m} \Big( 1- \frac{2}{2n+1}\Big)^{n (-1)^n} ,   
\end{aligned} 
\end{equation*}
and this completes the proof of formula $\eqref{catalani}$.  
Multiplying both sides of the latter formula by $e$ and using the fact that
$$ e = \lim_{m \to \infty} \Big(1 - \frac{2}{4m+3} \Big)^{-(2m+1)} $$ 
gives formula $\eqref{catalanii}$. Finally, multiplying formulas $\eqref{catalani}$ and $\eqref{catalanii}$ together and expanding gives formula $\eqref{catalaniii}$. 
 
\section{Proof of Proposition \ref{zeta3}}  

We will first prove formula $\eqref{zeta3i}$. 
\begin{lemma}\label{l31} 
Let $m \in {\mathbb N}$ and $\delta  \in  
\Big(0, \frac{1}{2}\Big)$. Then 
\begin{equation*}
\begin{aligned} 
&amp;\pi^2 \ \int_{\sigma=\delta}^{1/2} \ 
\Big(\frac{1}{2} \ - \ \sigma \Big) 
\ 
\frac{\operatorname{cos} 
\Big(\pi \big(2 m +  1\big) \sigma \Big) 
}{2 m  +  1} \ d \sigma   \\  
&amp;=    
\frac{\operatorname{cos} \Big( 
\pi  (2 m  +  1)  
 \delta \Big) 
}{\big(2 m  +  1\big)^3 
} \  
+ \ \frac{\pi  
\Big( \delta  -  \frac{1}{2} \Big) 
\  
\operatorname{sin} 
\Big( \pi  
\big(2  m  +  1\big) \delta \Big) 
}{\big(2 m  + 1\big)^2} . 
\end{aligned}
\end{equation*} 
\end{lemma} 
\begin{proof}
 This is straightforward 
integration by parts: 

\begin{equation*}
\begin{aligned} 
&amp;\int_{\sigma = \delta}^{1/2} 
\ \Big( \frac{1}{2} 
\ - \ \sigma \Big) \ 
\frac{\operatorname{cos} 
 \Big(\pi  \big(2  m  +  1\big)  \sigma 
 \Big) }{2  m  +  1} 
\  d  \sigma \\  
&amp;=  
\Bigg[  
\Big( \frac{1}{2} 
\ -  \sigma \Big)  
\frac{\operatorname{sin} 
\Big(\pi  \big(2  m  + 1\big)  \sigma 
\Big)}{\pi \big(2  m  +  1\big)^2} 
\Bigg]_{\sigma = \delta}^{1/2} 
\ + \    
\int_{\sigma = \delta}^{1/2} 
\ 
 \frac{ \operatorname{sin} 
\Big(\pi  \big(2  m  +  1\big)  \sigma 
\Big) }{\pi  \big(2  m  +  1\big)^2} 
\ d  \sigma \\  
&amp;=  
\Big( \delta - \frac{1}{2} \Big) \  
\frac{\operatorname{sin} 
\Big(\pi  \big(2 m  +  1\big)  \delta \Big) 
}{\pi  \big(2 m  +  1\big)^2 } \   
- \   
\Bigg[  
\frac{ \operatorname{cos} 
\Big(\pi  \big(2  m  +  1\big)  \sigma 
\Big) }{\pi^2  \big(2  m  +  1\big)^3} 
\Bigg]_{\sigma = \delta}^{1/2} \ ,  
\end{aligned}
\end{equation*}  
and the claim follows.  
\end{proof}

\begin{corollary}\label{c32}  
Let $m \in {\mathbb N}$.   
Then 
$$\pi^2 \  \int_{\sigma = 0}^{1/2} 
\  \Big(\frac{1}{2} 
 -  \sigma \Big) \  
\frac{\operatorname{cos} 
\Big( \pi  \big(2  m  +  1\big)  \sigma 
 \Big)}{2  m  +  1} 
\  d \sigma 
\ = \ 
\frac{1}{\big(2  m  +  1\big)^3}. 
$$

\end{corollary} 
\begin{proof}  
In Lemma $\ref{l31}$, let $\delta \to 0 +$. 
\end{proof}
We now recall the following basic formula: 
\begin{equation}\label{e31} 
\sum_{m = 0}^{\infty} 
\ \frac{1}{\big(2  m  +  1\big)^3} 
\ = \ \frac{7}{8} \ \zeta \big(3\big). 
\end{equation} 
We will establish the following: 

\begin{formula}\label{f33}  
$$\zeta \big(3\big) 
\ = \ 
\frac{4}{7} 
\ \pi \ G 
\  - \ 
\frac{2}{ 7} 
\ \pi^2 
\ \int_{\sigma = 0}^{1/2} 
\ \frac{\pi \sigma^2 
}{ \operatorname{sin} 
 \big(\pi  \sigma\big)} 
\ d \sigma. 
$$
\end{formula} 
\begin{proof}
First,  
we may rewrite Formula $\ref{f22}$ as 

\begin{equation}\label{e32}  
G \ = \ 
\int_{\sigma = 0}^{1/2} 
\  
\frac{\pi}{2} \  
\operatorname{log} \bigg( 
\operatorname{cot} \Big( 
\frac{\pi}{2} \sigma \Big)  
\bigg)  
\ d  \sigma . 
\end{equation} 
Second, by $\eqref{e31}$ and Corollary $\ref{c32}$, we have 

\begin{equation}\label{e33}  
\frac{7}{8} \ \zeta \big(3\big) 
 \ = \ 
\sum_{m = 0}^{\infty} 
\ \pi^2 \ \int_{\sigma = 0}^{1/2} 
\ \Big(\frac{1}{2} 
\ - \ \sigma \Big) \  
\frac{\operatorname{cos} 
\Big( \pi  \big(2  m  +  1\big)  \sigma 
\Big) }{2  m  +  1} 
\ d  \sigma .  
\end{equation} 
Fix $\delta  \in \Big( 0,  
\frac{1}{2} \Big)$.  
For each  $n  \in {\mathbb N}$,  
define the function 
$$
F_n \big(\sigma\big) 
\  
= \ \sum_{m=0}^n 
\ \Big( \frac{1}{2} \ - \ \sigma 
 \Big) \  
\frac{ 
\operatorname{cos} \Big( \pi (2 m  +  1)  \sigma  \Big) 
}{2  m  +  1} 
$$ 
on the interval 
$I \ = \  \Big[ \delta , \dfrac{1}{ 
2} \Big]$. 
The sequence 
$$
\Big\{ \sum_{m=0}^n 
\  
\operatorname{cos} \Big( \pi  \big(2  m  +  1\big)  
\sigma \Big)  \Big\}_{n \in {\mathbb N}} 
$$
of functions   
is uniformly bounded on $I$  
 by   
$( 2 \operatorname{sin} 
( \pi  \delta ) 
 )^{-1}$     
(see \cite[Formula (15), p.\ 198]{A} or 
\cite[Item 185.5, p.\ 316]{K}), whereas the sequence 
$$
\Big\{\Big( \frac{1}{2} 
\ - \ \sigma \Big) \  
\frac{1}{2 m  +  1 } 
\Big\}_{m \in {\mathbb N}}
$$ 
clearly tends monotonically to  $0$ uniformly on $I$. 
Hence by applying Dirichlet's test for 
uniform 
convergence (see \cite[Thm.~9.15, p. \ 230]{A} or  
or \cite[p.\ 347]{K}), 
it follows that 
the sequence 
of functions  
$F_n \big(\sigma\big)$  converges uniformly on $I$.  
Therefore, the series 
$$
\sum_{m=0}^{\infty}  
\  \Big(\frac{1}{2} \ - \ \sigma 
\Big) \  
\frac{ 
\operatorname{cos} \Big( \pi (2  m  +  1)  \sigma  \Big) 
}{2  m  +  1} 
$$
can be integrated term by term on  $I$.  
Hence, Lemma $\ref{l31}$ establishes the following     

\begin{formula}\label{f34}  
\begin{equation*}
\begin{aligned}
&amp;\pi^2 \  \int_{\sigma = \delta}^{1/2}  
\ \sum_{m=0}^{\infty}  
\ \Big( \frac{1}{2} \ - \ \sigma 
 \Big) \  
\frac{ 
\operatorname{cos} \Big( \pi (2 m  +  1)  \sigma  \Big) 
}{2  m  +  1} \ d  \sigma \\  
&amp;=   
\sum_{m = 0}^{\infty} \  
\frac{\operatorname{cos} \Big( 
\pi (2 m  +  1)  
 \delta \Big) 
}{\big(2  m  +  1\big)^3 
}  \ 
+ \  
\pi \  \Big( \delta \ - \ \frac{1}{2} \Big) 
\ 
\sum_{m = 0}^{\infty} \  
\frac{\operatorname{sin} 
\Big( \pi 
\big(2  m  + 1\big)  \delta  \Big) 
}{\big(2 \, m \, + \, 1\big)^2  }. 
\end{aligned}
\end{equation*}
\end{formula} 
Now take the limits of both sides of the latter formula as 
$\delta \to 0 +$. 
By the Weierstrass $M$-test, 
both 
series on the right-hand side of Formula $\ref{f34}$ are 
uniformly convergent series of functions of $\delta$  
on the 
interval  
$I = [ \delta  ,  \frac{1}{ 
2} ]$.  
Therefore, we can interchange limits and infinite sums
on the 
right-hand side of 
Formula $\ref{f34}$ (see \cite[Thm.~9.7, p.\ 220]{A}). 
By $\eqref{e33}$, it follows 
that 

\begin{equation}\label{e34}  
\pi^2 \  
\int_{\sigma=0}^{1/2} \  
\sum_{m=0}^{\infty} \  
\Big( \frac{1}{2} \ - \ 
\sigma \Big) 
\  \frac{\operatorname{cos}  
\Big( \pi (2 m  +  1)  \sigma  \Big) 
}{2  m  +  1} \ 
= \ \frac{7}{8} \ 
\zeta \big(3\big) .  
\end{equation} 
Combining $\eqref{e32}$, $\eqref{e34}$ and Formula $\ref{f21}$ gives 

\begin{equation*}
\begin{aligned}  
&amp;\frac{7}{8} \ 
\zeta \big(3\big) 
\ = \  
\pi^2 \ \int_{\sigma = 0}^{1/2} 
\ \Big( \frac{1}{2} 
\ -  \sigma \Big)  
 \ \frac{1}{2} 
\ \operatorname{log} 
\bigg(\operatorname{cot} 
\Big(\frac{\pi }{2} 
\sigma \Big) \bigg) \  
d \sigma 
\\ 
&amp;= \ \frac{\pi^2}{4} \ \Big(  
\int_{\sigma = 0}^{1/2} 
\ \operatorname{log} 
\bigg( \operatorname{cot} 
 \Big( \frac{\pi}{2} 
 \sigma \Big) \bigg)  
\ d \sigma 
\ - \ 
\int_{\sigma = 0}^{1/2} 
\ 2 \sigma  \operatorname{log} 
\bigg(\operatorname{cot} 
 \Big( \frac{ \pi }{2} 
 \sigma \Big) \bigg) \  
d  \sigma 
\Big) \\ 
&amp;=  \frac{\pi}{2} G 
\  - \  
\frac{ \pi^2}{2} 
\ \int_{\sigma = 0}^{1/2} 
\ \sigma \ \operatorname{log} 
\bigg(\operatorname{cot} 
 \Big( \frac{ \pi}{2} 
 \sigma  \Big) \bigg) \ 
d  \sigma .
\end{aligned}
\end{equation*}
In short, 

\begin{equation}\label{e35}  
\zeta \big(3\big) 
\ = \ 
\frac{4 }{7} \ \pi \ G 
\ - \  
\frac{4}{7} \ \pi^2 
\ \int_{\sigma = 0}^{1/2} 
\  \sigma \  \operatorname{log} 
\bigg( \operatorname{cot} 
 \Big( \frac{\pi}{2} 
 \sigma \Big) \bigg) \  
d \sigma .
\end{equation} 
Formula $\ref{f33}$ now follows because  

\begin{equation*}
\begin{aligned} 
&amp;\int_{\sigma = 0}^{1/2} 
\ \sigma \ \operatorname{log} 
\bigg(\operatorname{cot} 
\Big( \frac{\pi }{2} 
 \sigma  \Big) \bigg) \  
d  \sigma 
\\
&amp;=  
\Bigg[ 
\frac{\sigma^2 }{2} 
\  
\operatorname{log} 
\bigg( \operatorname{cot} 
 \Big( \frac{ \pi}{2} 
 \sigma  \Big) \bigg)  
\Bigg]_{\sigma = 0}^{1/2} 
\ - \ 
\int_{\sigma = 0}^{1/2} 
\  \frac{ \sigma^2 }{2} 
\ 
\frac{1}{ 
\operatorname{cot} 
 \Big( \frac{\pi}{2} 
\ \sigma \Big) } 
\  
\frac{-1}{ 
\Big(  \operatorname{sin} 
 \Big( \frac{ \pi}{2} 
 \sigma  \Big) \Big)^2 } 
\ \frac{ \pi }{2} \ d \, \sigma 
\\ 
&amp;=  0 \ + \ 
\int_{\sigma = 0}^{1/2} 
\ \frac{ \sigma^2 }{2} 
\ \frac{1}{ 
\operatorname{cos} 
 \Big( \frac{\pi}{2} 
 \sigma \Big) \  
\operatorname{sin} 
 \Big( \frac{\pi}{2} 
 \sigma  \Big) } 
\ \dfrac{\pi }{2} \  d  \sigma 
\ = \ \frac{1}{2} 
\ \int_{\sigma = 0}^{1/2} 
\ \frac{\pi  \sigma^2 
}{ \operatorname{sin} 
 \big(\pi  \sigma\big) } 
\ \sigma. 
\end{aligned}
\end{equation*}
\end{proof}
The following statement is similar to Lemma $\ref{l26}$. 

\begin{lemma}\label{l35}  
Let  $m \in {\mathbb Z}$, 
$m \geq  1$. Then 

$$
\int_{\sigma = 0}^{1/2} 
\ \frac{\sigma^3 }{ 
\sigma^2 \ - \ m^2} 
\  d  \sigma 
\ = \ 
\frac{1}{8} 
\ + \  
\frac{m^2 }{2} 
\ \operatorname{log} 
 \frac{ 4  m^2  -  1 
}{ 4  m^2}.
$$
\end{lemma} 
\begin{proof}
 This is straightforward: 

\begin{equation*}
\begin{aligned}
\int_{\sigma = 0}^{1/2} 
\ \frac{\sigma^3}{ 
\sigma^2 \ - \ m^2} 
\  d  \sigma 
&amp;=  
\int_{\sigma = 0}^{1/2} 
\  
\bigg( \sigma \ + \  
m^2 \ \frac{\sigma}{ \sigma^2 \ -  m^2 
}  \bigg) \  d  \sigma 
\\ 
&amp;=  
\Bigg[\  
\frac{1}{2} \sigma^2 
\  + \  m^2 \   
\frac{1}{ 2} \  
\operatorname{log}  
\Big( -  \sigma^2  +  m^2  \Big) 
\Bigg]_{\sigma = 0}^{1/2}\\  
&amp;=  
\frac{1}{8} 
\  + \  
\frac{ m^2}{2} 
\ \operatorname{log}  
\frac{ 
m^2 \ - \ \frac{1}{4} 
}{m^2} . 
\end{aligned}
\end{equation*}   
\end{proof} 

\begin{lemma}\label{l36} 
$$\lim_{n \to \infty} \frac{e^n}{\Big(1 +\frac{1}{n}\Big)^{n^2}} \ = \ \lim_{n \to \infty} \frac{e^{-n}}{\Big( 1 - \frac{1}{n} \Big)^{n^2}} \ = \ \sqrt{e} .  
$$  
\end{lemma} 
\begin{proof}  
By taking logarithms, it suffices to show that 


$$
\lim_{n \to \infty}  
\  
\Big(n \ - \ n^2 \ \operatorname{log}  
\Big(1 \ + \ \frac{1}{n} \Big)  
\Big) \  = \  
\dfrac{1}{2} 
\ = \  
\lim_{n \to \infty} 
\ 
\Big( - n \ - \ n^2 \ \operatorname{log}  
\Big(1 \ - \ \frac{1}{ n}  \Big)  
\Big) .
$$    
This follows by substituting  
$x \ = \ \pm  \frac{1}{n }$ 
in the 
Maclaurin series of   
the function $\operatorname{log}  \big(1 \ + \ x\big)$ 
and using continuity.   
\end{proof} 

We now proceed with the proof of 
formula $\eqref{zeta3i}$.  
By Corollary $\ref{c25}$, we have 
$$
\frac{\pi\, \sigma^2}{ 
\operatorname{sin} \big(\pi \sigma\big) 
} \  = \  
\sigma 
\ + \  
2 \sigma^3 \  
\sum_{m = 1}^{\infty} 
\ \frac{\big(-1\big)^m 
}{\sigma^2 
 -  m^2} . 
$$
Integrating both sides with respect to  $\sigma$ 
over the interval $\Big[ 0,  
\frac{1}{2} \Big]$ and using formula $\eqref{catalani}$ (and its proof) and Lemma $\ref{l35}$ gives  

\begin{equation*}
\begin{aligned}
&amp;\int_{\sigma = 0}^{1/2} 
\ 
\frac{\pi \sigma^2}{ 
\operatorname{sin}  \big( \pi  \sigma\big) 
} 
\  d  \sigma 
\ = \  
\int_{\sigma = 0}^{1/2} 
\  
\Big(  
\sigma 
\ + \  
2  \sigma^3 \  
\sum_{m = 1}^{\infty} 
\ \frac{\big(-1\big)^m 
}{\sigma^2 
 -  m^2 } 
\Big) \ d  \sigma 
\\ 
&amp;=   
\int_{\sigma = 0}^{1/2} 
\  
\sigma \ d  \sigma 
\ + \ 
2  
\int_{\sigma = 0}^{1/2} 
\  
\Big( \sum_{m = 1}^{\infty} 
\big(-1\big)^m \ 
\frac{\sigma^3 
}{\sigma^2 
 -  m^2 }  \Big) \  d  \sigma   
\\ 
&amp;=   
\frac{1}{8} 
\ + \ 
2  
\sum_{m = 1}^{\infty} 
\big(-1\big)^m \ 
\int_{\sigma = 0}^{1/2} 
\  
\frac{ \sigma^3 
}{ \sigma^2 
 -  m^2 } \ d  \sigma 
\\  
&amp;=  
\frac{1}{8} 
\ + \ 
2  
\sum_{m = 1}^{\infty} 
\big(-1\big)^m  \ 
\Big( 
\frac{1}{8} 
\ + \  
\frac{ m^2 }{2} 
\ \operatorname{log}  
\frac{ 
4 m^2  -  1 }{4  m^2} 
\Big) 
\\  
&amp;= 
\frac{1}{8 } 
\ + \ 
\sum_{m = 1}^{\infty} 
\big(-1\big)^m \ 
\Big( 
\frac{1}{4} 
\ + \  
m^2 \ \operatorname{log} \ 
\frac{ 
4  m^2  -  1 }{4  m^2} 
 \Big) ,  
\end{aligned} 
\end{equation*} 
which equals 

\begin{equation*}
\begin{aligned}  
&amp;\frac{1}{8} 
\ + \ 
\sum_{\ell = 1}^{\infty} 
\ \Big( -  \Big( 
\frac{1}{4 } 
\  + \ 
\big(2  \ell  -  1\big)^2 \ \operatorname{log} \ 
\frac{ 
4  \big(2  \ell  -  1\big)^2  -  1 
}{4  \big(2  \ell  -  1\big)^2}
 \Big)
\\ 
&amp;+   \Big( 
\frac{1}{4} 
\  + \  
\big(2 \ell\big)^2 \ \operatorname{log} \ 
\frac{ 
4 \big(2 \ell\big)^2  -  1 
}{4  \big(2  \ell\big)^2}
 \Big) \Big) \\   
&amp;=  
\frac{1}{8} 
\ + \ 
\sum_{\ell = 1}^{\infty} 
\  \Big( 
-  
\big(2  \ell  -  1\big)^2   \ \operatorname{log} \, 
\frac{ 
4  \big(2  \ell  -  1\big)^2  -  1 
}{4  \big(2  \ell  -  1\big)^2}
\\ 
&amp;+   
\big(2  \ell\big)^2 \  \operatorname{log} \ 
\frac{ 
4 \big(2  \ell\big)^2  -  1 }{4  \big(2  \ell\big)^2}
 \Big)\\  
&amp;=  
\frac{1}{8} 
\ + \ 
\sum_{\ell = 1}^{\infty} 
\ \operatorname{log} \ \Big(  
\frac{ 
\Big( 4  \ell  -  1  \Big)^{4 \ell -1} 
\  
\Big( 4  \ell  +  1  \Big)^{(2 \ell)^2} 
\  
\Big( 4  \ell  -  2  \Big)^{2 (2 \ell -1)^2} 
}{ \Big( 4  \ell  \Big)^{2 (2 \ell)^2} 
\  \Big( 4  \ell  -  3 \, \Big)^{(2\ell -1)^2} } 
 \Big) 
\\ 
&amp;=   
\dfrac{1}{ 8} 
\  + \  
\sum_{\ell = 1}^{\infty} 
\  \operatorname{log} \ \Big( 
\frac{\Big( 4  \ell  -  1 \Big)^{4 \ell -1} 
}{ 
\Big( 4  \ell  -  3  \Big)^{2 \ell -1} \  
\Big( 4  \ell  +  1  \Big)^{2 \ell}  } 
\Big) 
\\ 
&amp;+  \sum_{\ell = 1}^{\infty} \ 
\operatorname{log} \ \Big( 
\frac{ \Big( 4  \ell  +  1 \, \Big)^{4 \ell^2 + 2\ell} 
\  \Big( 4  \ell  -  2  \Big)^{2 (2\ell -1)^2}} 
{\Big( 4  \ell  \Big)^{2 (2 \ell)^2} \  
\Big( 4  \ell  -  3  \Big)^{4 \ell^2 - 6\ell +2}  }    
 \Big)  
\\ 
&amp;=   
\frac{2  G }{\pi} 
\  - \  \frac{3}{8} 
\  + \  2 \ 
\sum_{\ell = 1}^{\infty} \  
\operatorname{log} \ \Big(  
\frac{ \Big( 4  \ell  +  1  \Big)^{2\ell^2+\ell} 
\  \Big( 4  \ell  -  2 \, \Big)^{(2\ell-1)^2} 
}{ 
\Big( 4  \ell  \Big)^{(2\ell)^2} \  
\Big( 4  \ell  -  3  \Big)^{2\ell^2-3\ell+1} }   
\Big) . 
\end{aligned}
\end{equation*} 
Therefore, by Formula $\ref{f33}$, it follows that

\begin{equation}\label{e36}  
\dfrac{7}{4 \pi^2 } 
\ \zeta  \big(3\big) 
\  = \  
\dfrac{3}{ 16 } \ + \  
\operatorname{log} \  \prod_{\ell = 1}^{\infty}   \ 
\frac{ \Big( 4  \ell  \Big)^{(2\ell)^2}  \ 
\Big( 4  \ell  -  3  \Big)^{2\ell^2-3\ell+1} 
}{  
\Big( 4  \ell +  1  \Big)^{2 \ell^2 +\ell}  
\ \Big( 4  \ell  -  2 \Big)^{(2\ell-1)^2}  } \  .  
\end{equation} 
Now the latter infinite product can be written as   
$$\lim_{N\to \infty} \ \prod\limits_{\ell=1}^N \  2^{4 \ell -  1} \  
\frac{ \Big( 2 \ell  \Big)^{(2\ell)^2} 
}{ \Big(2  \ell  -  1\Big)^{(2\ell-1)^2} }
\  
\frac{ 
\Big( 4  \big(\ell  -  1\big)  +  1  
\Big)^{2(\ell-1)^2 + (\ell -1)} }{ 
\Big(4  \ell  +  1  \Big)^{2 \ell^2 + \ell} } ,  
$$
which equals 
  
\begin{equation*}
\begin{aligned} 
&amp;\lim_{N\to \infty}   
\bigg(\frac{2}{ 4  N  +  1 } 
\bigg)^{2 N^2  +  N} \  
\prod\limits_{\ell=1}^N \  
\frac{\Big( 2  \ell  \Big)^{(2\ell)^2} 
}{ \Big( 2  \ell  -  1 
 \Big)^{(2\ell-1)^2} } 
\\ 
&amp;=  \lim_{N\to \infty}  
\frac{2^{2N^2+N} \  \Big( 
2  N  +  1  \Big)^{(2N+1)^2} }{ 
\Big( 4  N  +  1  \Big)^{2 N^2 +N}} \   
\prod\limits_{\ell=1}^N \ 
\frac{ \Big( 2  \ell  \Big)^{(2\ell)^2} 
}{ 
\Big( 2  \ell  +  1  \Big)^{(2\ell+1)^2}  }  
\\ 
&amp;=  \lim_{N \to \infty}  
\Bigg( \frac{ 2^{2N^2+N} \  
\Big( 2  N  +  1  \Big)^{(2N+1)^2} }{ 
\Big(4  N  +  1  \Big)^{2N^2+N}} \  
\frac{2}{ 
\Big( 2  N  +  2  \Big)^{(2N+1)(N+1)}  } 
\\ 
&amp;\times    
\prod\limits_{\ell=1}^N \   
\frac{ \Big( 2  \ell  \Big)^{(2 \ell +1) \ell} 
\  
\Big( 2  \ell  +  2  \Big)^{(2\ell +1) (\ell +1)} 
}{  
\Big( 2  \ell  +  1  \Big)^{(2\ell+1)^2} } \Bigg) 
\\ 
&amp;= \lim_{N \to \infty}  
\Bigg(e^{\frac{{}_{N+1}}{{}^2}} 
\  \bigg( \frac{ 
4  N  +  2 }{ 
4  N  +  1 } \bigg)^{2N^2+N} \  
\bigg( \frac{ 2  N  +  1 
}{ 2  N  +  2 } 
 \bigg)^{(2N+1)(N+1)} \   
\frac{2}{\sqrt{e} } 
\\ 
&amp;  
\times  \prod\limits_{\ell=1}^N \  
\frac{\Big( 2  \ell  \Big)^{(2\ell+1) \ell} 
\  \Big( 2  \ell  +  2 \Big)^{(2\ell +1)(\ell+1)} 
}{ 
\sqrt{e} \  
\Big( 2  \ell  +  1 \Big)^{(2\ell+1)^2}} \Bigg).  
\end{aligned}
\end{equation*}
We claim that  

\begin{equation}\label{e37}  
\lim_{N \to \infty} 
\  
\frac{ e^{\frac{N+1}{2}}}{ 
\bigg( 1  
-  \frac{1}{4  N  +  2 }  
\bigg)^{(2N+1)N} \  
\bigg( 1  
+  \frac{1}{ 2  N  +  1 }  
\bigg)^{(2N+1)(N+1)} } \  = \  
e^{- \frac{3}{16}} .   
\end{equation} 
Indeed, by Lemma $\ref{l36}$, we have   

$$
\lim_{N \to \infty} \  
\frac{ e^{N + \frac{1}{4}}}{ 
\bigg( 1  +  
\frac{1}{2  N  +  1 } 
\bigg)^{\frac{(2N+1)^2}{2}}}  
\   = \ 1 \  = \  
\lim_{N \to \infty} \  
\frac{ e^{\frac{1}{2}}}{ 
\bigg( 1  +  
\frac{1}{2  N  +  1} 
\bigg)^{\frac{2N+1}{2}}}  
$$  
and 

$$ 
\lim_{N \to \infty} \  
\frac{e^{- \frac{N}{2} - 
\frac{5}{16}}}{ 
\bigg(1  -  \frac{1}{4  N  +  2 
}\Big)^{\frac{(4N+2)^2}{8}}} 
\ = \  1 \ = \  
\lim_{N \to \infty} \  
\frac{e^{\frac{1}{4}}}{ 
\bigg( 1 -  
\frac{1}{4 N  +  2 } 
\bigg)^{- \frac{4N+2}{4}}} \ ,  
$$    
hence $\eqref{e37}$ follows. Combining $\eqref{e36}$ with $\eqref{e37}$ gives 
$$ e^{\frac{7 \zeta(3)}{4 \pi^2}} = \frac{2}{\sqrt{e}} \ \prod_{l=1}^{\infty} 
\frac{(2l)^{(2l+1) l} \ (2l+2)^{(2l+2)(l+1)}}{\sqrt{e} \ (2l+1)^{2l+1)^2}} .$$  
Therefore, 
\begin{equation*} 
\begin{aligned} 
e^{\frac{7\zeta(3)}{4 \pi^2}} &amp;= \frac{2}{\sqrt{e}} \  
\lim_{m\to \infty} \prod\limits_{n=1}^{m} 
\frac{(2n)^{(2n+1)n} \ (2n+2)^{(2n+2)(n+1)}}{\sqrt{e} \ (2n+1)^{(2n+1)^2}} \\  
&amp;=\frac{(2m+2)^{(2m+1)(m+1)}}{e^{\frac{m+1}{2}}} \ \frac{2^{2^2} \cdot 4^{4^2} \cdot 6^{6^2} \cdots (2m)^{(2m)^2}}{3^{3^2} \cdot 5^{5^2} \cdot 7^{7^2} \cdots (2m+1)^{(2m+1)^2}}  \\ 
&amp;= e^{-\frac{1}{4}} \ \lim_{m\to \infty} \prod\limits_{n=1}^{2m+1} \frac{1}{\sqrt[4]{e}} \Big(1 - \frac{1}{n+1} \Big)^{\frac{n(n+1)}{2} (-1)^n} .  
\end{aligned}
\end{equation*} 
and this completes the proof of formula $\eqref{zeta3i}$. 

It remains to prove 
formulas $\eqref{zeta3ii}$ and $\eqref{zeta3iii}$. 
Note that 
$$\prod\limits_{n=1}^{2m} \sqrt[4]{e} \ \Big(1 - \frac{1}{n+1}\Big)^{\frac{n(n+1)}{2}(-1)^n} = \frac{\Big( 1 - \frac{1}{2m+2}\Big)^{(2m+1)(m+1)}}{e^{-(m+\frac{1}{4})}} \ \prod_{n=1}^{2m+1} \frac{1}{\sqrt[4]{e}} \ \Big( 1 - \frac{1}{n+1} \Big)^{\frac{n(n+1)}{2} (-1)^n} . $$   
Therefore, formula $\eqref{zeta3ii}$ will follow from formula $\eqref{zeta3i}$ 
once we show that 
$$ \lim_{m\to \infty} \frac{e^{-(2m+\frac{1}{2})}}{\Big(1 - \frac{1}{2m+2}\Big)^{(2m+2)(2m+1)}} = e . $$ 
This follows by writing $(2m+2)(2m+1)$ as $(2m+2)^2-(2m+2)$ and using 
Lemma $\ref{l36}$.   

Now multiplying formulas $\eqref{zeta3i}$ and $\eqref{zeta3ii}$ together and 
squaring gives

\begin{equation*} 
\begin{aligned} 
e^{\frac{7 \zeta(3)}{\pi^2}} &amp;= 
\lim_{m \to \infty} \ \frac{1}{\sqrt{e}} \  
\Big( \frac{2m+1}{2m+2} \Big)^{-(2m+1)(2m+2)} \prod\limits_{n=1}^{2m} \Big( 
1- \frac{1}{n+1} \Big)^{2n(n+1)(-1)^n}    \\ 
&amp;= \lim_{m \to \infty} \frac{1}{\sqrt{e}} \ \Big( \frac{(2m+2)^{2m+2}}{(2m+1)^{6m+2}} \Big)^{2m+1} \ \Big( \frac{2^{2^2} \cdot 4^{4^2} \cdot 6^{6^2} \cdots (2m)^{(2m)^2}}{1^{1^2} \cdot 3^{3^2} \cdot 5^{5^2} \cdots (2m-1)^{(2m-1)^2}}  \Big)^4     
\end{aligned}
\end{equation*} 
Formula $\eqref{zeta3iii}$ is now a consequence of the equality  
$$ \frac{1}{\sqrt{e}} = \lim_{m\to \infty} 
\Big( \frac{2m+1}{2m+2} \Big)^{m+2} .  $$ 

\section{Proof of Facts \ref{zeta3gs} and \ref{catalangs}} 

By formula $\eqref{zeta3iii}$ and its proof, it suffices to show that the 
total 
exponent of $k+1$ in 
the 
infinite product expansion given by Guillera and Sondow 
\cite[Example~5.3]{GS}  
equals  
$\big(-1\big)^{k+1} \ \big(k  +  1\big)^2$, 
for all $k  \in {\mathbb N}$. 
The exponent in question 
equals 

\begin{equation*}
\begin{aligned} 
&amp;\big(-1\big)^{k+1} \  
\sum_{n=k}^{\infty} \  
{n \choose k} \  
\frac{n^2  +  n }{2^{n+3} } 
\ =  \ \frac{ 
\big(-1\big)^{k+1} }{8 \ \cdot 
\big(k  !  \big) 
} 
\ \sum_{n=k}^{\infty} \ 
\frac{ \big(n  +  1\big) \ n^2 
\ \big(n  -  1\big) \ \cdots \ 
\big(n  -  k  +  1\big)}{ 2^n} 
\\    
&amp;=   
\frac{ 
\big(-1\big)^{k+1}}{8 \ \cdot 
\big( k  !  \big) } 
\  
\Big( \sum_{n=k}^{\infty} \ 
\frac{ 
\big(n+2\big) \ \big(n+1\big) \ \cdots \ \big(n-k+1\big)}{2^n} 
-  2  
\sum_{n=k}^{\infty} \ 
\frac{\big(n+1\big) \ n \ \cdots \ \big(n-k+1\big)}{2^n}  \Big)  
\\  
&amp;=  \frac{ 
\big(-1\big)^{k+1} }{ 8 \ \cdot 
\big( k  !  \big) } 
\  
\Big(  
\frac{1}{ 2^k} 
\ \sum_{m=0}^{\infty} \  
\frac{\big(m+k+2\big) \ \big(m+k+1\big) \ \cdots \ 
\big(m+1\big) \ }{ 2^m } 
\\ 
&amp;-   
\frac{1}{ 2^{k-1} } \  
\sum_{m=0}^{\infty} \  
\frac{\big(m+k+1\big) \ \big(m+k\big) \ \cdots \ 
\big(m+1\big)}{ 2^m }  \Big) .
\end{aligned}
\end{equation*}
We have the following lemma: 

\begin{lemma}\label{l41}  
For all  
$k  \in {\mathbb N}$, we 
have  

$$
\sum_{m=0}^{\infty} \  
\frac{ \big(m+k+1\big) \ \big(m+k\big) \ \cdots \ 
\big(m+1\big) }{  2^m } \
 = \  
2^{(k+2)} \ \cdot \ \big( \big(k+1\big) !  \big).
$$     
\end{lemma} 
\begin{proof} 
This follows by term-by-term $\big(k+1\big)$-fold 
differentiation of the geometric series 
$$
\sum_{n=0}^{\infty} \ x^n \  = \ (1 -  x)^{-1}
$$  
and subsequent evaluation at 
$x  =  \frac{1}{2}$.  
\end{proof}

Therefore, by Lemma $\ref{l41}$, the exponent in question equals 
$$ 
\frac{ 
(-1)^{k+1}}{8 \ \cdot 
\ (k ! ) } 
\  
\big(8 \ \cdot \ \big( \big(k+2\big) !  \big) 
\ - \ 8 \ \cdot \ \big( \big(k+1\big) !  \big)  \big) 
\ = \ (-1)^{k+1} \ 
(k +  1)^2 , 
$$ 
which completes the proof of Fact $\ref{zeta3gs}$. 

We will now show that, apart from the factor $e^{-\frac{1}{2}}$ on the left-hand side of formula $\eqref{catalani}$, 
the product expansion given by the latter formula and the product 
expansion given by Guillera and Sondow 
\cite[Example~5.5]{GS} are equivalent. In other words,  
we will show that the total 
exponent of $2  k  +  1$ in the 
infinite product expansion of  
$e^{\frac{G}{\pi}}$ 
given by Guillera and Sondow 
\cite[Example~5.5]{GS} equals  
$(-1)^{k+1} \ 
( k +  \frac{1}{2} )$, 
for all 
$k  \in {\mathbb N}$. Since the infinite series involved  
is only conditionally convergent, the discrepancy involving 
$e^{-\frac{1}{2}}$ can be explained   
by means of Riemann's theorem 
on rearrangements of conditionally convergent series.  
The exponent in question equals 

\begin{equation*}
\begin{aligned} 
&amp;\big(-1\big)^{k+1} \  
\sum_{n=k}^{\infty} \  
{n \choose k} \  
\frac{n}{ 2^{n+2}} 
\ = \  
\frac{\big(-1\big)^{k+1} 
}{ 4 \ \cdot \ 
\big( k  !  \big) } \  
 \sum_{n=k}^{\infty} \  
\frac{n^2 \  
\big(n  -  1\big) \ \cdots \ 
\big(n  -  k  +  1\big) 
}{ 2^n } 
\\ 
&amp;= 
\frac{ 
\big(-1\big)^{k+1} }{ 4 \ \cdot \ 
\big( k  !  \big) } \ 
\Big( 
\sum_{n=k}^{\infty} \  
\frac{ \big(n+1\big) \ n \ \cdots \ 
\big(n-k+1\big) }{ 2^n } 
\  - \  
\sum_{n=k}^{\infty} \  
\frac{ n \ \big(n-1\big) 
\ \cdots \ \big(n-k+1\big) }{ 2^n } 
\Big) 
\\  
&amp;= 
\frac{ 
\big(-1\big)^{k+1}  }{4 \ \cdot \ 
\big( k  !  \big) } \ 
\Big( \frac{1}{ 2^k } \  
\sum_{m=0}^{\infty} 
\frac{ \big(m+k+1\big) \ \big(m+k\big) \ \cdots \ 
\big(m+1\big) \ }{ 2^m } 
\\ 
&amp;-   
\frac{1}{ 2^k } \ 
\sum_{m=0}^{\infty} \  
\frac{\big(m+k\big) \ \big(m+k-1\big) 
\ \cdots \ \big(m+1\big) 
}{ 2^m }   \Big) . 
\end{aligned}
\end{equation*}
By Lemma $\ref{l41}$, this equals 
$$ \frac{ (-1)^{k+1} 
}{ 4 \ \cdot \ 
\big(k  !  \big) } 
\  \big( 4  \ \cdot \ 
\big( \big(k  +  1\big)  !  \big) 
\ - \ 2 \ \cdot \ \big( k  !  \big) 
\big) \  
= \ 
\big(-1\big)^{k+1} \ 
\big( k  +  \frac{1}{ 2} \big) \ , $$  
as required,   
and this completes the proof of Fact $\ref{catalangs}$. 

\section{Concluding remarks} 

\begin{remark}
 The identities 
$$ \sum_{n=k}^{\infty} {n \choose k} \frac{n^2+n}{2^{n+3}} = (k+1)^2 \ , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\sum_{n=k}^{\infty} {n \choose k} \frac{n}{2^{n+2}} = k + \frac{1}{2} $$ 
which were used in the proofs of Facts $\ref{zeta3gs}$ and $\ref{catalangs}$ 
can also be very easily established
by the Wilf-Zeilberger method via the use of Zeilberger's Maple package EKHAD  
(see \cite{PWZ}).   
\end{remark} 

\begin{remark}
One way to account for the fact that the products discussed in this paper 
are so closely
tied to the ones studied by Guillera and Sondow in \cite{GS} is by 
noticing that they are related via  
Euler transformations. For instance, 
using the latter formula in the previous remark, one has
$$ \lim_{m\to \infty} \sum_{k=1}^{2m} (-1)^{k} k \operatorname{log} \frac{2k-1}{2k+1} = \lim_{m\to \infty} \sum_{k=1}^{2m} (-1)^k \operatorname{log} \frac{2k-1}{2k+1} \sum_{n=k}^{\infty} {n \choose k} \frac{n-1}{2^{n+2}} .$$  
If we interchange the summation  
on the right-hand side (an Euler 
transformation) the relation between formula $\eqref{catalani}$ and the formula given in Fact $\ref{catalangs}$ becomes evident.    
\end{remark}

\begin{remark} 
The formulas in Propositions $\ref{catalan}$ and $\ref{zeta3}$ are 
reminiscent of some 
powerful statements that deserve to be 
more widely known. We refer the reader to Finch's book \cite{F} 
for a wealth of 
information regarding such statements involving classical constants.  
For instance, the following function (first introduced by Borwein 
and Dykshoorn in \cite{BD}):  
$$D(x) = \lim_{m\to \infty} \prod_{n=1}^{2m+1} \big(1+ \frac{x}{n}\big)^{n (-1)^{n+1}} = e^x \lim_{m\to \infty} \prod_{n=1}^{2m} \big(1 +\frac{x}{n} \big)^{n (-1)^{n+1}} . $$ 
Certain values of this function are related to some classical constants. 
Melzak proved in \cite{M} that $D(2) = \frac{\pi e}{2}$.  
In \cite{BD},  
Borwein and Dykshoorn generalized Melzak's result and 
explicitly determined 
the values of $D(x)$ at all rational $x$ having denominator 1, 2 or 3. 
Interestingly enough, some of
the  
resulting evaluations involve Catalan's constant, the 
Glaisher-Kinkelin constant and $\Gamma(\frac{1}{4})$. We have not been able 
to show that any of the formulas in Propositions $\ref{catalan}$ or  
$\ref{zeta3}$ is a direct consequence of the latter 
evaluations. 
\end{remark}

\section{Acknowledgment} 

We are much obliged to the referee for providing valuable references and insightful remarks on a previous version of this paper.
    
\begin{thebibliography}{9}

\bibitem{A}
T. Apostol, {\it Mathematical Analysis},  
Addison Wesley, 1974. 

\bibitem{BB}
J. Borwein and D. Bailey, {\it Mathematics by Experiment}, A. K. Peters, 2004.  

\bibitem{BD} 
P. Borwein and W. Dykshoorn, An interesting infinite product, {\it J. Math. 
Anal. Appl.}, {\bf 179} (1993), 203--207. 

\bibitem{F} 
S. R. Finch, {\it Mathematical Constants}, Cambridge University Press, 2003. 
  
\bibitem{GS}
J. Guillera and J. Sondow, Double integrals and infinite products for some 
classical constants via analytic continuations of Lerch's transcendent, {\it 
Ramanujan J.} {\bf 16} (2008), 243--270.  

\bibitem{K}
K. Knopp, {\it Theory and Application of Infinite Series}, Dover, 1990. 

\bibitem{M} 
Z. A. Melzak, Infinite products for $\pi e$ and $\pi/e$, {\it Amer. Math. Monthly}, {\bf 68} (1961), 39--41. 

\bibitem{N}
P. Nahin, {\it Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills}, 
Princeton University Press, 2006.  

\bibitem{PWZ} 
M. Petk{\u o}vsek, H. Wilf, and D. Zeilberger, {\it $A=B$}, A. K. Peters, 1996.

\end{thebibliography}


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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11Y60; Secondary 11M06, 33B99, 40A25.

\noindent \emph{Keywords: }
Catalan's constant, zeta(3), infinite product formula.  

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A002117} and 
\seqnum{A006752}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 1 2012;
revised version received November 13 2012.  
Published in {\it Journal of Integer Sequences}, November 20 2012.

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