Product under Summation (Part 1)

Let's just rewrite it in a way to make it slightly more obvious what we need to do.

$S= 1 + \frac{1}{2^2} + \frac{1}{4^2}\left(1+\frac{1}{2^2} \right) + \frac{1}{6^2}\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{4^2} \right) + \frac{1}{8^2} \left(1+\frac{1}{2^2}\right) \left(1+\frac{1}{4^2}\right) \left(1+\frac{1}{6^2} \right) + \ldots$

Oh look, we can factorise $1+\frac{1}{2^2}$ .

$S=\left( 1 + \frac{1}{2^2}\right) \left[1+ \frac{1}{4^2} + \frac{1}{6^2}\left(1+\frac{1}{4^2} \right) + \frac{1}{8^2} \left(1+\frac{1}{4^2}\right) \left(1+\frac{1}{6^2} \right) + \ldots\right]$

Wait, now we can factorise $1+\frac{1}{4^2}$ .

$S=\left( 1 + \frac{1}{2^2}\right) \left(1+ \frac{1}{4^2}\right)\left[1 + \frac{1}{6^2}+ \frac{1}{8^2} \left(1+\frac{1}{6^2} \right) + \ldots\right]$

Now we can factorise $1+\frac{1}{6^2}$ ! So if we keep doing it we eventually get:

$S=\prod_{n=1}^{\infty} \left(1+\frac{1}{(2n)^2}\right)$

Using the fact that $\large \frac{\sinh(x)}{x}=\prod\limits_{n=1}^{\infty} \left(1+\frac{x^2}{\pi^{2}n^{2}}\right)$ and by letting $\large x=\frac{\pi}{2}$ we get:

$\large S=\frac{\sinh\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}}$

---

Can anyone rigorously show that the infinite series and the infinite product are identical?

It's just a telescoping series! $\prod_{n=1}^{N+1} \left(1 + \frac{1}{4n^2}\right) - \prod_{n=1}^N \left(1 + \frac{1}{4n^2}\right) \; = \; \frac{1}{4(N+1)^2}\prod_{n=1}^N \left(1 + \frac{1}{4n^2}\right)$ and so $\sum_{N=1}^M \frac{1}{4(N+1)^2}\prod_{n=1}^N \left(1 + \frac{1}{4n^2}\right) \; = \; \prod_{n=1}^{M+1} \left(1 + \frac{1}{4n^2}\right) - \frac54 \;,$ which proves, letting $M \to \infty$ , that $\frac54 + \sum_{N=1}^\infty \frac{1}{4(N+1)^2}\prod_{n=1}^N \left(1 + \frac{1}{4n^2}\right) \; = \; \prod_{n=1}^\infty \left(1 + \frac{1}{4n^2}\right) \; = \; \frac{2}{\pi}\sinh\big(\tfrac12\pi\big) \;.$