TITLE: Evaluation of $\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$
QUESTION [15 upvotes]: During my calculations I ended up with the following product:
$$P=\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$$
I tried to express it in terms of series by taking the logarithm 
$$S=\ln P=\sum_{n=1}^\infty \ln\left(e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\right)$$
but I also got stuck. Numerical calculation suggests that it is equal to
$$P\stackrel{?}=\frac{\sqrt{2\pi}}{e}$$
but I am not able to prove the conjecture. Any idea about how to evaluate the product? Any help would be appreciated. Thanks in advance.

REPLY [9 votes]: We have:
$$\log P = \sum_{n=1}^{+\infty}\left(1+\left(n+\frac{1}{2}\right)\log\left(1-\frac{1}{n+1}\right)\right)$$
but:
$$\sum_{n=1}^{N}\left(1+n\log n-(n+1)\log(n+1)\right) = N-(N+1)\log(N+1)$$
since we have a telescopic sum, while:
$$\sum_{n=1}^{N}\left(\frac{1}{2}\log n+\frac{1}{2}\log(n+1)\right)=\log(N!)+\frac{1}{2}\log(N+1)$$
so:
$$\sum_{n=1}^{N}\left(1+\left(n+\frac{1}{2}\right)\log\left(1-\frac{1}{n+1}\right)\right)=N+\log(N!)-\left(N+\frac{1}{2}\right)\log(N+1)$$
and the result follows from the Stirling approximation:
$$\log(N!) = \left(N+\frac{1}{2}\right)\log N-N+\log\sqrt{2\pi}+O\left(\frac{1}{N}\right).$$