How do I find:
$$\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right) \quad$$
I am pretty sure that the infinite product converges, but if it doesn't please let me know if I have made an error.
Also, could I have a nice explanation as to how would someone arrive to the answer.
Thanks alot.
Best Answer
An infinite product $\prod_{n \ge 0} (1 + u_n)$ where $u_n > 0$ converges if and only if $\sum_{n \ge 0} u_n$ converges, and $\sum_{n \ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}$ is a famous result by Euler.
Euler's product formula for $\sin z$ is hard to prove, but intuitive (it has roots at $\pm n \pi$): $$ \frac{\sin z}{z} = \prod_{n \ge 1} \left( 1 - \frac{z^2}{\pi^2 n^2} \right) $$ This gives directly $\dfrac{\sin i}{i} = \dfrac{e - e^{-1}}{2}$ for your product