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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.

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

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  • $\begingroup$ I like this one but the the last term should be $\displaystyle\frac{\sqrt{2\pi}}{e}$ $\endgroup$ – Venus Nov 25 '14 at 12:28
  • $\begingroup$ Very nice. I guess you mean $\sqrt{2\pi}$ in the last line. $\endgroup$ – Simon S Nov 25 '14 at 12:28
  • $\begingroup$ Yes, thanks for pointing that out! Typo fixed. Thanks for the nice comments and the upvotes! $\endgroup$ – hypergeometric Nov 25 '14 at 12:31
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    $\begingroup$ Nice. The $\approx$ should be $=$ though. $\endgroup$ – Antonio Vargas Nov 25 '14 at 15:43
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    $\begingroup$ And why not? :) $\endgroup$ – hypergeometric Jan 23 '15 at 16:32
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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: $$\log(N!) = \left(N+\frac{1}{2}\right)\log N-N+\log\sqrt{2\pi}+O\left(\frac{1}{N}\right).$$

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