Calculate $\sum_{n\ge2}\log\left(1-\frac1{n^2}\right)$

This is the expression whose sum I have to calculate:

$$\sum_{n\ge2}\log\left(1-\frac1{n^2}\right)$$

I have tried to use the mengoli's series properties but I failed. The listed answer should be $$-\log (2)$$.

• It is also easy to show by telescopic product or by induction that $$\prod_{n=2}^N\,\left(1-\frac{1}{n^2}\right)=\frac{N+1}{2N}$$ for every integer $N\geq 2$. – Batominovski Oct 13 '18 at 16:07
Hint: $$\log\left(1-\frac{1}{n^2}\right)=\log\left(\frac{n^2-1}{n^2}\right)=\log\left(\frac{(n-1)(n+1)}{n\cdot n}\right)\\=\log(n+1)+\log(n-1)-2\log(n)=\left(\log(n+1)-\log(n)\right)-\left(\log(n)-\log(n-1)\right)$$ now use telescoping sum method.