Prove that $\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$ I friend of mine sent me this problem a while ago, but I still can't figure it out. (He can't figure it out either.) I figured here would be a good place to ask for help.

Prove:
  $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$

 A: $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
$$\log\left(\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \log\left(\frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(\log({e^{-2}}) + n\log\left(\frac{n+1}{n-1}\right)\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log\left(\frac{n+1}{n-1}\right) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log(n+1) - n\log(n-1) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(2n\coth^{-1}(n) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$2\sum_{n=2}^\infty \left(n\coth^{-1}(n) - 1\right) = 3 - \log(4\pi)$$
Here I get stuck however.
A: The logarithm of the LHS equals
$$ L=\sum_{n\geq 2}\left[-2+2n\,\text{arctanh}\frac{1}{n}\right]=\sum_{n\geq 2}\sum_{m\geq 1}\frac{1}{\left(m+\frac{1}{2}\right)n^{2m}}=\sum_{m\geq 1}\frac{\zeta(2m)-1}{m+\frac{1}{2}}\tag{1} $$
and we may recall that
$$ \frac{1-\pi z\cot(\pi z)}{2} = \sum_{m\geq 1}\zeta(2m) z^{2m} \tag{2} $$
to get that:
$$ L = \int_{0}^{1}\left(\frac{1-3z^2}{1-z^2}-\pi z\cot(\pi z)\right)\,dz\tag{3} $$
and $L=3-\log(4\pi)$ follows from the connection between the primitive of $z\cot(z)$ and the dilogarithm, proving the stated conjecture: 
$$ \int z\cot(z)\,dz = z\log\left(1-e^{2iz}\right)-\frac{i}{2}\left(z^2+\text{Li}_2(e^{2iz})\right)\tag{4} $$
Nice question!
A: Consider the partial product
$$P_N = \prod_{n=2}^{N} \frac1{e^2} \left (\frac{n+1}{n-1} \right )^n $$
By writing out the terms of $P_N$, we can easily find a simple expression for it:
$$P_N = e^{-2(N-1)} \frac{N^{N-1} (N+1)^N}{2 (N-1)!^2} $$
The result follows by using Stirling's approximation as well as the definition of $e$ as $N \to \infty$:
$$(N-1)! \approx \sqrt{2 \pi (N-1)} (N-1)^{N-1} e^{-(N-1)} $$
$$ (N+1)^N \approx e N^n $$
A: First, recombining terms gives
$$
\begin{align}
\prod_{n=2}^m\color{#C00}{\frac{1}{e^2}}\left(\frac{\color{#090}{n+1}}{\color{#00F}{n-1}}\right)^n
&=\color{#C00}{\frac1{e^{2(m-1)}}}\color{#090}{\prod_{n=3}^{m+1}n^{n-1}}\color{#00F}{\prod_{n=1}^{m-1}n^{-n-1}}\\
&=\frac1{e^{2(m-1)}}\left(\frac{(m+1)^m}2\prod_{n=1}^mn^{n-1}\right)\left(m^{m+1}\prod_{n=1}^mn^{-n-1}\right)\\
&=\frac1{e^{2(m-1)}}\frac{(m+1)^mm^{m+1}}{2m!^2}
\end{align}
$$
Then, applying Stirling, we get
$$
\begin{align}
\lim_{m\to\infty}\frac1{e^{2(m-1)}}\frac{(m+1)^mm^{m+1}}{2m!^2}
&=\lim_{m\to\infty}\frac1{e^{2m-3}}\frac{m^{2m+1}e^{2m}}{4\pi m^{2m+1}}\\
&=\frac{e^3}{4\pi}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mc{P}_{N} \equiv
\prod_{n = 2}^{N}\bracks{{1 \over \expo{2}}\pars{n + 1 \over n-1}^{n}}\,,\qquad \lim_{N \to \infty}\mc{P}_{N} = {\expo{3} \over 4\pi}:\ {\large ?}}$

$$
\begin{array}{c}
{\large\mathsf{I'll\ show\ that\ \,\mc{P}_{N}\ has\ a\ closed\ expression.}}
\\
\bbox[15px,#ffe,border:1px dotted navy]{\large%
\mbox{A full and}\ \underline{detailed\ derivation}\ \mbox{is given as follows:}}
\end{array}
$$
\begin{align}
\ln\pars{\mc{P}_{N}} & =
\sum_{n = 2}^{N}\bracks{-2 + n\,\ln\pars{n + 1 \over n - 1}} =
\sum_{n = 2}^{N}\bracks{%
-2\int_{0}^{1}\dd t + n\int_{0}^{1}{2\,\dd t \over 2t + n - 1}}
\\[5mm] & =
2\int_{0}^{1}\pars{1 - 2t}\sum_{n = 0}^{N - 2}{1 \over n + 2t + 1}\,\dd t
\\[5mm] & =
2\int_{0}^{1}\pars{1 - 2t}
\sum_{n = 0}^{\infty}\pars{{1 \over n + 2t + 1} - {1 \over n + N + 2t}}\,\dd t
\\[5mm] & =\int_{0}^{1}\pars{1 - 2t}
\partiald{}{t}\ln\pars{\Gamma\pars{2t + N} \over \Gamma\pars{2t + 1}}\,\dd t
\\[5mm] & =
-\ln\pars{\Gamma\pars{2 + N} \over \Gamma\pars{3}} -
\ln\pars{\Gamma\pars{N} \over \Gamma\pars{1}} +
\int_{0}^{2}\ln\pars{\Gamma\pars{t + N} \over \Gamma\pars{t + 1}}\,\dd t
\\[1cm] & =
-\ln\pars{\bracks{N + 1}\Gamma\pars{N + 1}} + \ln\pars{2} -
\ln\pars{\Gamma\pars{N + 1} \over N + 1}
\\[3mm] & \phantom{=\,\,\,}+
\int_{0}^{2}\ln\pars{\Gamma\pars{t + N} \over \Gamma\pars{t + 1}}\,\dd t
\\[1cm] & =
\int_{0}^{2}\ln\pars{\Gamma\pars{t + N} \over \Gamma\pars{t + 1}}\,\dd t -
2\ln\pars{N!} + \ln\pars{2}\label{1}\tag{1}
\end{align}

However,

\begin{align}
&\partiald{}{\alpha}\int_{0}^{2}\ln\pars{\Gamma\pars{t + \alpha}}\,\dd t =
\int_{0}^{2}\partiald{\ln\pars{\Gamma\pars{t + \alpha}}}{t}\,\dd t =
\ln\pars{\Gamma\pars{\alpha + 2} \over \Gamma\pars{\alpha}} =
\ln\pars{\bracks{\alpha + 1}\alpha}
\\[5mm] & \implies
\int_{0}^{2}\ln\pars{\Gamma\pars{t + N} \over \Gamma\pars{t + 1}}\,\dd t =
\int_{1}^{N}\ln\pars{\bracks{\alpha + 1}\alpha}\,\dd\alpha
\\[5mm] & =
2 - 2N + \ln\pars{N + 1} - 2\ln\pars{2} + N\ln\pars{N} + N\ln\pars{N + 1}
\label{2}\tag{2}
\end{align}

\eqref{1} and \eqref{2} yield

\begin{align}
\ln\pars{\mc{P}_{N}} & =
2 - 2N + \ln\pars{N + 1} - \ln\pars{2} + N\ln\pars{N} + N\ln\pars{N + 1} -2\ln\pars{N!}
\\[1cm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,\require{cancel}
2 - \cancel{2N} + \bracks{\cancel{\ln\pars{N}} + \ln\pars{1 + {1 \over N}}} - \ln\pars{2} + \cancel{N\ln\pars{N}}
\\[3mm] & +
\bracks{\cancel{N\ln\pars{N}} + N\ln\pars{1 + {1 \over N}}} -
2\bracks{{1 \over 2}\ln\pars{2\pi} + \cancel{\pars{N + {1 \over 2}}\ln\pars{N}} - \cancel{N}}
\\[1cm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
2 +\ \underbrace{N\ln\pars{1 + {1 \over N}}}
_{\ds{\to\ 1\ \,\mrm{as}\ N\ \to\ \infty}}\ -\ \ln\pars{4\pi}
\implies
\bbx{\lim_{N \to \infty}\mc{P}_{N} = \ln\pars{\mrm{e}^{3} \over 4\pi}}
\end{align}

$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\prod_{n = 2}^{\infty}\bracks{{1 \over \expo{2}}\pars{n + 1 \over n-1}^{n}} =
{\expo{}^{3} \over 4\pi}}}
$$
