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?

  • $\begingroup$ Is it convergent? I got the product to be $$\frac {3^5 \cdot 5^9 \cdot 7^{13} \cdots} {2^3 \cdot 4^7 \cdot 6^{11} \cdots}$$ which is definitely divergent. $\endgroup$ – Dbchatto67 Mar 14 at 7:32
  • $\begingroup$ There are two limits possible, please see my answer. $\endgroup$ – user90369 Mar 14 at 16:51

$\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:

$$A=2^{1/36}\pi^{1/6}\left( \lim\limits_{N\to\infty} \frac{e^{N/2}N^{-1/8}}{ \prod\limits_{n=1}^{N}\left(1+\frac{1}{2n}\right)^n }\right)^{2/3}\left( \lim\limits_{N\to\infty} \frac{e^{2N}(2N)^{-1/2}}{ \prod\limits_{n=1}^{2N}\left(1+\frac{1}{n}\right)^n }\right)^{-1/3}$$

(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:

$\displaystyle \lim\limits_{N\to\infty}\prod\limits_{n=1}^{2N}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n} = 2^{1/6}\pi^{1/2}eA^{-6} \approx 1.2157517513…$

$\displaystyle \lim\limits_{N\to\infty}\prod\limits_{n=1}^{2N+1}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n} = 2^{1/6}\pi^{1/2}A^{-6} \approx 0.44725…$

  • $\begingroup$ thank you for showing the full answer $\endgroup$ – user550260 Mar 15 at 20:31

Let $$a_n=\left(\frac{n}{n+1}\right)^{(-1)^{n-1} n}$$ then $$a_{2p}= \left(\frac{2p}{2 p+1}\right)^{-2 p}\qquad \text{and}\qquad a_{2p+1}=\left(\frac{2 p+1}{2 p+2}\right)^{2 p+1}$$ Now, using a CAS, $$\prod_{p=1}^m a_{2p}=\frac{\sqrt[12]{2} \sqrt{\pi } \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+2 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)+\frac{1}{4}\right)}{A^3 \,\Gamma \left(m+\frac{3}{2}\right)}$$ $$\prod_{p=1}^m a_{2p+1}=\frac{2 \sqrt[12]{2} \Gamma (m+2) \exp \left(2 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta ^{(1,0)}(-1,m+2)+\frac{1}{4}\right)}{A^3}$$ $$b_m=\frac 12\prod_{p=1}^m a_{2p}\prod_{p=1}^m a_{2p+1}$$ $$b_m=\frac{2^{\frac 16}\sqrt{\pi } \Gamma (m+2) \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+4 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta ^{(1,0)}(-1,m+2)+\frac{1}{2}\right)}{A^6 \,\Gamma \left(m+\frac{3}{2}\right)}$$ $$b_m=\frac{2^{\frac 16} \sqrt{\pi }\, \Gamma (m+2)}{A^4 \,H(m)^2\,\Gamma \left(m+\frac{3}{2}\right)}\exp \left(4 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta ^{(1,0)}(-1,m+2)+\frac{1}{3}\right)$$ where appears the hyperfactorial function.

Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^{\log(b_m)}$

$$b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}\left(1+\frac{1}{8 m}-\frac{49}{384 m^2}+\frac{127}{1024 m^3}+O\left(\frac{1}{m^4}\right) \right)$$

$$\color{blue}{\lim_{m\to \infty } \, b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}}$$

  • $\begingroup$ Thank you for the answer @Claude Leibovic $\endgroup$ – user550260 Mar 14 at 10:16
  • $\begingroup$ @coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result. $\endgroup$ – Claude Leibovici Mar 14 at 10:22
  • $\begingroup$ @coffeee. Take care : I could be wrong by a factor of two since I used $b_m=\prod_{p=1}^m a_{2p}\prod_{p=1}^m a_{2p+1}$ instead of $b_m=\prod_{p=1}^m a_{2p}\prod_{p=\color{red}{0}}^m a_{2p+1}$. I need to check again tomorrow morning. $\endgroup$ – Claude Leibovici Mar 14 at 14:42

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