Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$ 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.

$$ \sum_{n=2}^{\infty} \ln \left( 1 - \frac{1}{n^2} \right) = \ln \left( \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right) \right) $$
$$ = \ln \left( \prod_{n=2}^{\infty} \frac{n-1}{n}\frac{n+1}{n} \right) $$
$$ \ln \left( \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times ... \right) $$
$$ = \ln \left( \frac{1}{2} \right) = -\ln 2 $$
