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

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.

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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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