Answer
$-\ln 2$
Work Step by Step
The nth term of the series can be rewritten as:
$\ln (1-\frac{1}{n^2})=\ln (\frac{n^2-1}{n^2})=\ln (\frac{n+1}{n}\cdot \frac{n-1}{n})=\ln \frac{n+1}{n}-\ln \frac{n}{n-1}=\ln (1+1/n)-\ln (1+1/(n-1))$
Notice that:
$\sum_{n=2}^\infty\ln (1+\frac{1}{n})-\ln (1+\frac{1}{n-1})$
$=(\ln (1+1/2)-\ln (1+1))+(\ln (1+1/3)-\ln (1+1/2))+(\ln (1+1/4)-\ln (1+1/3))+\ldots$
$=-\ln (1+1)$
$=-\ln 2$
Thus, $\sum_{n=2}^\infty \ln (1-\frac{1}{n^2})=-\ln 2$
