${\displaystyle \prod_{n=1}^{\infty} ({ \frac n {n+1}}) ^{(-1)^n}}$

Typed this infinite product into Wolfram Alpha and I got an approximate result of 1.5708. I wonder if anyone studied this infinite product because I couldn't find anything about it on the web and thought that it was quite interesting. I'd be pleased if anyone can inform me about this infinite product.


We can combine terms of the form $2n$ and $2n+1$. So, the product becomes $$\prod_{n=1}^\infty \bigg(\frac{2n}{2n-1}\bigg)\bigg(\frac{2n}{2n+1}\bigg)$$Which equals$$\prod_{n=1}^\infty \frac{4n^2}{4n^2-1}=\prod_{n=1}^\infty 1+\frac1{4n^2-1}$$This can be written as $$\lim_{n\to\infty}\frac1{2n+1}\cdot\frac{2^{4n}(n!)^4}{(2n)!^2}$$Which, by Stirling's approximation becomes $\frac\pi2$.

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    $\begingroup$ Thanks for the explanation! $\endgroup$ – B.Aytekin Sep 8 '18 at 20:38
  • $\begingroup$ Just a small comment on the language: you are combining two factors, not terms! A sum is made up of terms and a product is made up of factors. $\endgroup$ – Kavi Rama Murthy Sep 8 '18 at 23:34
  • $\begingroup$ @KaviRamaMurthy I believe that both are valid in the case of infinite products, but correct me if I'm wrong $\endgroup$ – Don Thousand Sep 9 '18 at 0:07

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