How to prove $ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$ How can I prove that
$$ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$
The result is given here (result 48).
The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.
Some of my attempts include:

*

*Multiplying known infinite products for $\frac{\pi}{2}$  and
$\frac{1}{e}$
$$ \frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots} $$
$$ \frac{1}{e} =  \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$
as well as others given in https://arxiv.org/pdf/1005.2712.pdf


*Trying to find the partial products from ${n=1} $ to $k $, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}} $$
However, I have been unsuccessful producing partial products on my own.


*Taking the $\ln$ of the product to try to find the partial sums;


*Trying to transform the following integral into infinite product.
$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e} $$
From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.
Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.
Thank you kindly for your help and time.
 A: The stated product does not converge and therefor the answer is incorrect.
It's only valid if the final value is even.
$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \neq  \frac{\pi}{2e}$$
Most likely you ment:
$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} =  \frac{\pi}{2e}$$
For the proof of even value's or the product above:
Consider:
$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} =  $$
$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} \big(n \ln(1+2/n)-2\big)\bigg)$$
We know the factorial (e.g.  stirlings formula, but people don't seem to write it this way, derived by writing it as sums, alternatively you can use divergent products, see: Divergent products.)
$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2 n\pi}\big) \exp\bigg(-\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg) $$
$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} 2\ln\bigg(\frac{(1+2/n)^{n/2})}{e}\bigg)=$$
$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2-1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n-1/2})}{e}\bigg)=$$
$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n})}{e}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$
$$\exp\bigg(2\sum_{n=1}^{\infty} \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1)^j}-\frac{\zeta(-j)}{j(n)^j}\bigg)-\bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1/2)^j}-\frac{\zeta(-j)}{j(n-1/2)^j}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$
if $\Delta_n^+=f(n)-f(n+1)$ and if f(n)= 0 as n goes to infty.
$$\exp\bigg(2\sum_{n=1}^{\infty} \sum_{j=1}^\infty \Delta_n^+\frac{\zeta(-j)}{j(n)^j}-\Delta_n^+\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n-1/2)^j}- \Delta_n^+\frac{1}{2}\ln\big(\frac{n-1/2}{n}\big)\bigg)=$$
$$\exp 2 \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j}-\sum_{j=1}^\infty \frac{\zeta(-j)}{j(1/2)^j}+\frac{\ln(2)}{2}\bigg)=$$
$$\exp 2 \bigg(\ln\big(\frac{\sqrt{2\pi}}{e}\big)-\ln(\frac{\sqrt{e}}{2\sqrt{2}})+\frac{\ln(2)}{2}\bigg)=$$
$$\frac{2\pi}{e^2}\frac{e}{8} 2=\frac{\pi}{2e}$$
A: This answer is spliced from the paper linked in comments
Melzak (1961) shows that the largest-volume cylinder (Cartesian product of a hypersphere and a line) in an $n$-dimensional unit sphere occupies this proportion of the sphere:
$$\rho_n=\frac{2(n\pi)^{-1/2}(1-1/n)^{(n-1)/2}\Gamma(n/2+1)}{\Gamma((n+1)/2)}$$
We have $\rho_2=\frac2\pi$ and $\lim_{n\to\infty}\rho_n=\sqrt{\frac2{\pi e}}$. Now define
$$\sigma_n=\frac{\rho_{n+2}}{\rho_n}=\sqrt{\left(\frac n{n+2}\right)^n\left(\frac{n+1}{n-1}\right)^{n-1}}$$
Telescoping on $\sigma_n$ for $n=2,4,6\dots$ then gives
$$\sqrt{\frac\pi{2e}}=\prod_{n=1}^\infty\left(\frac n{n+1}\right)^n\left(\frac{2n+1}{2n-1}\right)^{(2n-1)/2}$$
Squaring gives
$$\frac\pi{2e}=\prod_{n=1}^\infty\left(\frac {2n}{2n+2}\right)^{2n}\left(\frac{2n+1}{2n-1}\right)^{2n-1}=\prod_{n=1}^\infty(1+2/n)^{(-1)^{n+1}n}$$
A: Maybe try splitting it into two products to get rid of the alternating sign:
$$\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}n}=\prod \left(1+\frac2{2n}\right)^{-2n}\prod\left(1+\frac2{2n+1}\right)^{2n+1}$$
then look at these individually, the first one:
$$P_1=\prod_{n=1}^\infty\left(1+\frac1n\right)^{-2n}=\left[\prod_{n=1}^\infty\left(1+\frac1n\right)^n\right]^{-2}$$
although this diverges you may be able to get it to cancel with something from the other expression?
A: So far these are just some thoughts.

Introduction:
We can use the product $\sin\pi x=\pi x\prod_{n\ge1}(1-x^2/n^2)$ to show that
$$\mathrm L(x)=\int_0^x\ln\sin\pi t\,dt=\ln\left[\left(\frac{\pi x}{e}\right)^x\prod_{n\ge1}e_k(x)\right],\tag1$$
where
$$e_k(x)=\frac{j(n+x)}{(en)^{2x}j(n-x)}, $$
and $j(x)=x^x$. This is done by expanding
$$\ln\sin\pi t=\ln(\pi t)+\sum_{n\ge1}\ln(1-t^2/n^2)$$
and integrating termwise. On the other hand, we know a lot about a closely related function called the Clausen function, which is defined as
$$\mathrm{Cl}_2(x)=-\int_0^x\ln\left|2\sin\tfrac{t}{2}\right|dt=\sum_{n\ge1}\frac{\sin nx}{n^2}.\tag2$$
We can show that
$$\tau(x)=\prod_{n\ge1}e_n(x)=\left(\frac{e}{\pi x}\right)^x\exp\mathrm L(x)=\left(\frac{e}{2\pi x}\right)^x\exp\left[-\frac{1}{2\pi}\mathrm{Cl}_2(2\pi x)\right].\tag3$$

Main Ideas:
Consider the value of $\tau(1)$. We have
$$\mathrm{Cl}_2(2\pi)=\sum_{n\ge1}\frac{\sin2\pi n}{n^2}=0,$$
so $$\tau(1)=\frac{e}{2\pi}.$$
We will consider the convergent analog of the divergent product in the title
$$P=\prod_{n\ge1}\left(\frac{n}{n+1}\right)^{2n}\left(\frac{2n+1}{2n-1}\right)^{2n-1}.$$
We aim to show that $P=\frac{1}{4\tau(1)}=\frac{\pi}{2e}$.
From the Clausen duplication formula and a bit of algebraic manipulation, we may show that
$$\tau(2x)=\frac12\left(\frac{\pi}{e}\right)^{2x-1}\left(\frac{\tau(x)}{j(\tfrac12-x)\tau(\tfrac12-x)}\right)^2.\tag4$$
If we can find a suitable value $x$ to plug into $(4)$, we may be able to get an expression for $\frac{1}{4\tau(1)}$ which can be transformed, via algebraic manipulations, to the expression for $P$. I will update my answer once a suitable value is found.
