TITLE: Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$
QUESTION [13 upvotes]: So I have this problem for midterm reviews:
$$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$
I know that you can find the series form of a natural log, as shown here:
$$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$
But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.

REPLY [2 votes]: An overkill is as follows $$\pi^2\frac{\sin x} {x(\pi^2-x^2)} =\prod_{n=2}^{\infty} \left(1-\frac {x^2}{n^2\pi^2}\right) $$ Taking limits as $x\to\pi$ and then applying logarithm we get the desired result.