How to establish the convergence of this infinite product?

Question

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

Answer 1

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 $\sum_{n=1}^{+\infty} a_n < +\infty$

Answer 2

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