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  $-\frac{1}{2}$.

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.
