# How does this infinite product work?

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

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 $\frac\pi2$.