So I stumbled on this question by trying different infinite products on my calculator and trying to find interesting ones that do converge. I calculated this specific product to some larger n and found that it is equal (to enough decimal places to call it equal) to:
$$\prod_{k=1}^{\infty} \left(1+\frac{1}{k^2}\right) \approx 3.67607791… \approx \frac{1}{2\pi}(e^\pi-e^{-\pi})$$
I found this equality by searching the decimal number online, to my surprise I didn't find any mention of this product. I only got lucky and found this number as the result of the following integral:
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^tdt$$
Now I tried showing that these are equal algebraically, however I failed spectacularly at that. Here is what I tried:
$$\frac{1}{2\pi}(e^x-e^{-x})=\frac{1}{2\pi}\lim \limits_{n \to \infty}\left(1+\frac{\pi}{n}\right)^n-\left(1-\frac{\pi}{n}\right)^n=\frac{1}{2\pi}\lim \limits_{n \to \infty}\frac{(n+\pi)^n-(n-\pi)^n}{n^n}$$
This is where I stopped because I don't know what to do with this $(a+b)^n-(a-b)^n$ term or if this is even the right approach.
I also tried to write the initial product as:
$$\prod_{k=1}^{\infty} \left(1+\frac{1}{k^2}\right)=\prod_{k=1}^{\infty} \frac{k^2+1}{k^2}=\prod_{k=1}^{\infty} \frac{(k+i)(k-i)}{k^2}$$
However contrary to my hopes I wasn't able to do anything with this, other than this I don't really know how to solve these infinite products.
Is it possible to show this equality and if so how?
PS: Sorry if this question is a duplicate, it does feel like it, but I did not find anything about this simple product anywhere.
Best Answer
There is a standard infinite product representation for the sine function: $$ \sin x = x\prod_{k \geq 1} \left(1 - \frac{x^2}{(\pi k)^2}\right). $$ This holds for all $x \in \mathbf C$. Setting $x = i\pi$, $$ \sin(i\pi) = i\pi \prod_{k \geq 1} \left(1 + \frac{1}{k^2}\right). $$ From $\sin(t) = (e^{it} - e^{-it})/(2i)$, we get $\sin(i\pi)/(i\pi) = (e^{-\pi} - e^\pi)/(-2\pi) = (e^\pi - e^{-\pi})/(2\pi)$.