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?
 A: 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 
$$\sum_{n=2}^\infty \log\left(1-\frac{1}{n^2}\right)=-\log(2)$$
A: Hint. One may observe that
$$
\prod_{n=2}^N\left(1-\frac1{n^2}\right)=\prod_{n=2}^N\frac{n^2-1}{n^2}=\prod_{n=2}^N\frac{n+1}{n}\cdot \prod_{n=2}^N\frac{n-1}{n}
$$ then factors telescope.
Take $\ln$ of it and make $N \to \infty$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{n = 2}^{\infty}\ln\pars{1 - {1 \over n^{2}}}} & =
\sum_{n = 2}^{\infty}\bracks{%
\ln\pars{1 - {1 \over n}} + \ln\pars{1 + {1 \over n}}}
=
\sum_{n = 2}^{\infty}\,\,\int_{0}^{1}\pars{%
-\,{1 \over n - t} + {1 \over n + t}}\,\dd t
\\[5mm] = &
\int_{0}^{1}\sum_{n = 0}^{\infty}\pars{%
{1 \over n + 2 + t} - {1 \over n + 2- t}}\,\dd t
\\[5mm] & =
\int_{0}^{1}\bracks{\Psi\pars{2 - t} - \Psi\pars{2 + t}}\,\dd t
\qquad\qquad\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & =
\left.\vphantom{\LARGE A}%
-\ln\pars{\Gamma\pars{2 - t}\Gamma\pars{2 + t}}\,\right\vert_{\ 0}^{\ 1}
\qquad\qquad\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & =
-\ln\pars{\Gamma\pars{1}\Gamma\pars{3}} + \ln\pars{\Gamma\pars{2}\Gamma\pars{2}}
=
-\ln\pars{0! \times 2!} + \ln\pars{1! \times 1!}
\\[5mm] & =
\color{#f00}{-\ln\pars{2}}
\end{align}
