calculus - Evaluating the sum $sum_{n=2}^{infty}ln[1-1/n^2]$

For convenience the sum is again
$$
\sum_{n=2}^{\infty}\ln[1-1/n^2]=\sum_{n=2}^{\infty}\ln\frac{(n^2-1)}{(n^2)}
$$
I first tried solving using a definite integral, since this seems to make telescoping easier to see,
$$
\sum_{n=2}^{\infty}\ln\frac{(n^2-1)}{(n^2)}=\sum_{n=2}^{\infty}\ln(n^2-1)-\ln(n^2)=
\sum_{n=2}^{\infty}\int_{n^2}^{n^2-1}\frac{dx}{x}
$$
But writing out the first few terms doesn't make any cancellation obvious because of the $n^2$ terms.

I also tried futzing around with log rules and got things down to
$$
\sum_{n=2}^{\infty}\ln\frac{(n^2-1)}{(n^2)}=
\sum_{n=2}^{\infty}\ln((n-1)(n+1))-\ln(n^2)=
\sum_{n=2}^{\infty}\ln(n-1)+\ln(n+1)-\ln(n^2)
$$
The first few terms of which are
$$
\sum_{n=2}^{4}\ln\frac{(n^2-1)}{(n^2)}=[\ln1+\ln3-2\ln 2]+[\ln2+\ln4-2\ln 3]+[\ln3+\ln5-2\ln 4]+...\\

=\ln 2+\ln 5-2\ln 4
$$
Which leads to the guess that
$$
\sum_{n=2}^{4}\ln\frac{(n^2-1)}{(n^2)}=\ln (N-1)+\ln(N+2)-2\ln(N)\\
=\ln(\frac{(N-1)(N+2)}{N^2})\rightarrow 0
$$
Which means I'm wrong. Should I soldiering on looking for a pattern through more computation, or is there a more expedient/elegant way to evaluate the sum?

Answer

You were on the right track in writing $$\log\left(1-\frac{1}{n^2}\right)=\log(n+1)-\log(n)+\log(n-1)-\log(n)$$

Proceeding, we can write

$$\begin{align}
\sum_{n=2}^N \log\left(1-\frac{1}{n^2}\right)&=\sum_{n=2}^N \left(\log(n+1)-\log(n)\right)+\sum_{n=2}^N \left(\log(n-1)-\log(n)\right)\\\\
&=\log(N+1)-\log(2)-\log(N)
\end{align}$$

Therefore, we find that

$$\sum_{n=2}^\infty \log\left(1-\frac{1}{n^2}\right)=-\log(2)$$