How to establish the convergence of this infinite product? How to prove that $\prod\limits_{n=1}^\infty \cos^2(\frac{1}{n^2})$ is convergent ?
I have got the discussion here but could not figure out if we are asked to prove the convergence, through simple way how can we achieve that?
Thanks in advance
 A: If you only want convergence(for an infinite product of positive numbers smaller than $1$ it means telling whether the product is equal to $0$), here is a simple way:
$$\prod_{n=1}^\infty \cos^2(\frac{1}{n^2}) = \prod_{n=1}^\infty(1- \sin^2(\frac{1}{n^2}))$$.
$\sum_{n=1}^\infty \sin^2(\dfrac{1}{n^2}) < \infty$ since $\sin^2{\frac{1}{n^2}}\sim \frac{1}{n^4}$
So we know that $\prod_{n=1}^\infty \cos^2(\frac{1}{n^2}) > 0$.
I used the fact: If all $a_n \in (0,1)$, $\displaystyle\prod_{n=1}^{+\infty} (1- a_n)$ is non-zero if and only if $\sum_{n=1}^{+\infty} a_n < +\infty$
A: I would use two things :
First : $1-cos^2(\frac{1}{n^2})$ ~ $\frac{1}{n^4} $ , when n->$+\infty$
And : ln(1+u) ~ u , when u->0
If you let $P_n$ be your product from 1 to n, consider ln($P_n$):
$ln(P_n) = \sum_{k=1}^n ln(cos^2(\frac{1}{k^2})) = \sum_{k=1}^n ln(1-(1-cos^2(\frac{1}{k^2}))) $
$ln(1-(1-cos^2(\frac{1}{k^2}))) = -\frac{1}{n^4} +o(\frac{1}{n^4})$
So the series converges, i.e $ln(P_n)$ converges, and $P_n$ does as well
