Evaluation of $\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$ 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.
 A: 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:
$$\log(N!) = \left(N+\frac{1}{2}\right)\log N-N+\log\sqrt{2\pi}+O\left(\frac{1}{N}\right).$$
A: $$\begin{align}
P
&=\large\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\
&=\large\lim_{m\to\infty}\prod_{n=1}^m e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\
&=\large\lim_{m\to\infty}e^m\left[\left(\frac 12\right)^1\left(\frac 23\right)^2\left(\frac 34\right)^3\cdots \left(\frac m{m+1}\right)^m\right]
\sqrt{\frac12\cdot\frac23\cdot\frac34\cdots\frac{m}{m+1}}\\
&\large=\lim_{m\to\infty}e^m\left[\frac{1\cdot 2\cdot 3\cdots m}{(m+1)^m}\color{blue}{\cdot \frac{m+1}{m+1}}\right]\sqrt\frac1{m+1}\\
&=\large\lim_{m\to\infty}e^m\frac{\color{green}{(m+1)!}}{(m+1)^{m+3/2}}\\
&\large= \lim_{m\to\infty}\frac{e^m}{(m+1)^{m+3/2}}\color{green}{\left[\sqrt{2\pi(m+1)}\left(\frac{m+1}e\right)^{m+1}\right]\quad \text{(Stirling's approx)}}\\
&\large= \frac{\sqrt{2\pi}}e\qquad \blacksquare
\end{align}$$
