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

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

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

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 $N \to \infty$.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}