Sum the log!

We have:

Same solution as Abdelghani ßoris presented differently.

$\begin{aligned} \sum_{k=2}^\infty \ln\left(1 - \frac{1}{k^2} \right) & = \sum_{k=2}^\infty \ln\left(\frac{k^2 - 1}{k^2} \right) \\ & = \sum_{k=2}^\infty \ln\left(\frac{(k-1)(k+1)}{k^2} \right) \\ & = \lim_{n \to \infty}\ln \left( \prod_{k=2}^n \frac{(k-1)(k+1)}{k^2} \right) \\ & = \lim_{n \to \infty} \ln \left( \frac{\displaystyle \prod_{k=\color{#D61F06}{1}}^{\color{#D61F06}{n-1}}k \prod_{k=\color{#D61F06}{3}}^{\color{#D61F06}{n+1}}k}{\displaystyle \prod_{k=\color{#D61F06}{2}}^{\color{#D61F06}{n}}k^2} \right) \\ & = \lim_{n \to \infty} \ln \left( \frac{n+1}{2n} \right) \\ & = \lim_{n \to \infty} \ln \left( \frac{1+\frac{1}{n}}{2} \right) \\ & = \ln \left( \frac{1}{2} \right) = \boxed{-0.693} \end{aligned}$