I am looking for the closed form of this product.
$$\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^{7/6}$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
Best Answer
$\displaystyle 1/\prod\limits_{n=1}^{2N}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n} = \frac{1}{\sqrt{2}} \left( \frac{e^{N/2}N^{-1/8}}{ \prod\limits_{n=1}^{N}\left(1+\frac{1}{2n}\right)^n } \right)^4 \left( \frac{e^{2N}(2N)^{-1/2}}{ \prod\limits_{n=1}^{2N}\left(1+\frac{1}{n}\right)^n } \right)^{-1} $
$\displaystyle \lim\limits_{N\to\infty} \frac{e^{2N}(2N)^{-1/2}}{ \prod\limits_{n=1}^{2N}\left(1+\frac{1}{n}\right)^n } = \lim\limits_{N\to\infty}\frac{e^N N^{-1/2}}{ \prod\limits_{n=1}^N\left(1+\frac{1}{n}\right)^n } = \frac{\sqrt{2\pi}}{e}\enspace\enspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $\frac{\sqrt{2\pi}}{e}$ and the right with it’s product. After a few simple elementary conversions follows:
$\displaystyle \lim\limits_{N\to\infty}\prod\limits_{n=1}^{2N}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n} = 2^{1/6}\pi^{1/2}eA^{-6} \approx 1.2157517513…$
$\displaystyle \lim\limits_{N\to\infty}\prod\limits_{n=1}^{2N+1}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n} = 2^{1/6}\pi^{1/2}A^{-6} \approx 0.44725…$