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.
 A: $$\begin{align*}\log(1-1/n^2)&=\log\left(\frac{n^2-1}{n^2}\right)\\&=\log(n^2-1)-\log(n^2)\\&=\log[(n+1)(n-1)]-2\log(n)\\&=\log(n+1)+\log(n-1)-2\log(n)\end{align*}$$... and our series telescopes. The only term that survives is $-\log 2$.$$\begin{align*}\sum_{n=2}^\infty\log(1-1/n^2)&=\sum_{n=2}^\infty\left(\log(n+1)+\log(n-1)-2\log (n)\right)\\&=\log3+\log1\color{red}{-2\log2}+\log4+\color{red}{\log2}-2\log3+\dots\\&=-\log 2\end{align*}$$
A: $$ \sum_{n=2}^{\infty} \ln \left( 1 - \frac{1}{n^2} \right) = \ln \left( \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right) \right) $$
$$ = \ln \left( \prod_{n=2}^{\infty} \frac{n-1}{n}\frac{n+1}{n} \right) $$
$$ \ln \left( \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times ... \right) $$
$$ = \ln \left( \frac{1}{2} \right) = -\ln 2 $$
A: 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. 
A: By telescoping
setting  $u_n =\ln\left(\frac{n}{n-1}\right)$,we have $$\ln\left(1-\frac{1}{n^2}\right)=\ln\left(1+\frac{1}{n}\right)+\ln\left(1-\frac{1}{n}\right) =\ln\left(\frac{n+1}{n}\right)-\ln\left(\frac{n}{n-1}\right) $$that is 
$$\color{blue}{\ln\left(1-\frac{1}{n^2}\right)=u_{n+1}-u_n}$$

Whence,
  $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=  \sum_{n=2}^\infty (u_{n+1}-u_n) = \lim_{n\to\infty}u_n-u_2 = -u_2 =-\ln 2. $$

