Find Sum of $\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$. Prove that it converges. Question : For $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$$
a. Prove it converges
b. Find the sum

My Try
$
= \sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)\\
= \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 - \frac{1}{16})\\
= \ln(\frac{3}{4}) + \ln (\frac{8}{9}) + \ln ( \frac{15}{16})\\
= -.287 + -.117 + -.064 = -.50
$
it converges to  $-\frac{1}{2}$.
 A: Hint
$$\ln\left(1-\frac1{n^2}\right)=\ln\left(\frac{(n-1)(n+1)}{n^2}\right)=\ln\left(\frac{n-1}{n}\right)-\ln\left(\frac{n}{n+1}\right)=u_n-u_{n+1}$$
and then telescope. 
A: Hint:
Rewrite it as:
$$\sum_{n=2}^{\infty} ln(\frac{n^{2} - 1}{n^{2}}) = \sum_{n=2}^{\infty} ln( \frac{(n-1)(n+1)}{n^{2}}) = \sum_{n=2}^{\infty} [ ln(n-1) + ln(n+1) - 2ln(n)]$$
So we have: $ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...$
Think about this as a telescoping series.
A: Notice that
$$
a_n=\ln\left(1-\frac{1}{n^2}\right)<0 \quad \forall n \ge 2
$$Since the series
$$
\sum_{n=1}^\infty b_n:=\sum_{n=1}^\infty\frac{1}{n^2}
$$
is convergent, and 
$$
\lim_{n\to\infty}\frac{-a_n}{b_n}=\lim_{x\to0}\frac{-\ln(1-x)}{x}=1> 0,
$$
thanks to the Limit Comparison Test the series $\sum_{n=2}^\infty-a_n=-\sum_{n=2}^\infty a_n$ is convergent, and so is the series $\sum_{n=2}^\infty a_n$.
Furthermore, for every $n\ge 2$ we have
\begin{eqnarray}
s_n:&=&\sum_{k=2}^n\ln\left(1-\frac{1}{k^2}\right)=\sum_{k=2}^n\left[\ln(k^2-1)-\ln k^2\right]=\sum_{k=2}^n\left[\ln(k-1)+\ln(k+1)-2\ln k\right]\\
&=&\sum_{k=2}^{n-1}\ln k+\sum_{k=3}^{n+1}\ln k-2\sum_{k=2}^n\ln k=-\ln2+\ln n+\ln(n+1)-2\ln n\\
&=&-\ln2+\ln\left(1+\frac1n\right).
\end{eqnarray}
Since
$$
\lim_{n\to\infty}s_n=\lim_{n\to\infty}\left[-\ln2+\ln\left(1+\frac1n\right)\right]=-\ln 2,
$$
it follows that $\displaystyle \sum_{n=2}^\infty\ln\left(1-\frac{1}{n^2}\right)=-\ln2$
