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: $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
On
$$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
$$\log\left(\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \log\left(\frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(\log({e^{-2}}) + n\log\left(\frac{n+1}{n-1}\right)\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log\left(\frac{n+1}{n-1}\right) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log(n+1) - n\log(n-1) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(2n\coth^{-1}(n) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$2\sum_{n=2}^\infty \left(n\coth^{-1}(n) - 1\right) = 3 - \log(4\pi)$$
Here I get stuck however.
On
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 $N \to \infty$:
$$(N-1)! \approx \sqrt{2 \pi (N-1)} (N-1)^{N-1} e^{-(N-1)} $$
$$ (N+1)^N \approx e N^n $$
On
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 $$ \begin{align} \lim_{m\to\infty}\frac1{e^{2(m-1)}}\frac{(m+1)^mm^{m+1}}{2m!^2} &=\lim_{m\to\infty}\frac1{e^{2m-3}}\frac{m^{2m+1}e^{2m}}{4\pi m^{2m+1}}\\ &=\frac{e^3}{4\pi} \end{align} $$
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mc{P}_{N} \equiv \prod_{n = 2}^{N}\bracks{{1 \over \expo{2}}\pars{n + 1 \over n-1}^{n}}\,,\qquad \lim_{N \to \infty}\mc{P}_{N} = {\expo{3} \over 4\pi}:\ {\large ?}}$
$$ \begin{array}{c} {\large\mathsf{I'll\ show\ that\ \,\mc{P}_{N}\ has\ a\ closed\ expression.}} \\ \bbox[15px,#ffe,border:1px dotted navy]{\large% \mbox{A full and}\ \underline{detailed\ derivation}\ \mbox{is given as follows:}} \end{array} $$ \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}
On
This is too long for a comment.
Starting from robjohn's answer, it is interesting to notice that we can approximate quite accurately the partial products$$P_m=\prod_{n=2}^m\frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n$$ Using Stirling approximation with more terms $$P_m=\frac{e^3}{4 \pi }\left(1-\frac{2}{3 m}+\frac{5}{9 m^2}-\frac{209}{405 m^3}+O\left(\frac{1}{m^4}\right)\right)$$ $$\left( \begin{array}{ccc} m & \text{exact} & \text{approximation} \\ 5 & 1.415232961 & 1.414162459 \\ 10 & 1.499926976 & 1.499854086 \\ 15 & 1.531035192 & 1.531020352 \\ 20 & 1.547199306 & 1.547194537 \\ 25 & 1.557103354 & 1.557101382 \\ 30 & 1.563794263 & 1.563793306 \\ 35 & 1.568617521 & 1.568617002 \\ 40 & 1.572259374 & 1.572259069 \\ 45 & 1.575106537 & 1.575106346 \\ 50 & 1.577393544 & 1.577393418 \end{array} \right)$$
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!