TITLE: Prove that $\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$
QUESTION [31 upvotes]: 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:
  $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$

REPLY [3 votes]: 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
$$P_m=\frac{e^3}{4 \pi }\left(1-\frac{2}{3 m}+\frac{5}{9 m^2}-\frac{209}{405
   m^3}+O\left(\frac{1}{m^4}\right)\right)$$
$$\left(
\begin{array}{ccc}
m & \text{exact} & \text{approximation} \\
 5 & 1.415232961 & 1.414162459 \\
 10 & 1.499926976 & 1.499854086 \\
 15 & 1.531035192 & 1.531020352 \\
 20 & 1.547199306 & 1.547194537 \\
 25 & 1.557103354 & 1.557101382 \\
 30 & 1.563794263 & 1.563793306 \\
 35 & 1.568617521 & 1.568617002 \\
 40 & 1.572259374 & 1.572259069 \\
 45 & 1.575106537 & 1.575106346 \\
 50 & 1.577393544 & 1.577393418
\end{array}
\right)$$