Proving the convergence of $\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$

$$log(1-\frac{1}{n^2})= \\log(\frac{n^2-1}{n^2})=\\log(\frac{n-1}{n})+log(\frac{n+1}{n})\\s_{1}+s_{2}\\s_{1}=log(\frac{2-1}{2})+log(\frac{3-1}{3})+log(\frac{4-1}{4})+...++log(\frac{n-1}{n})=\\+log(\frac{1}{2}\frac{2}{3}\frac{3}{4}...\frac{n-1}{n})\\=log(\frac{1}{n})\\$$ $$ s_{2}=log(\frac{2+1}{2})+log(\frac{3+1}{3})+log(\frac{4+1}{4})+...++log(\frac{n+1}{n})=\\log(\frac{3}{2}\frac{4}{3}\frac{5}{4}...\frac{n+1}{n})\\=log(\frac{n+1}{2})\\s_{1}+s_{2}=\\log(\frac{1}{n})+log(\frac{n+1}{2})=log(\frac{n+1}{2n}) \rightarrow log(\frac{1}{2}) $$

$$\sum_{n=2}^\infty\ln (1-1/n^2)=\ln\prod_{n=2}^\infty\frac{n^2-1}{n^2}=\ln\lim_{N\to\infty}\prod_{n=2}^N\frac{n-1}n\prod_{n=2}^N\frac{n+1}n$$ Now note that $$\frac12\cdot\frac23\cdot\ldots\frac{n+1}n=\frac1n\\ \frac32\cdot\frac43\cdot\ldots\frac n{n+1}=\frac n2$$

Therefore the answer is $$\ln\frac12=-\ln2$$

$$\sum_{n=2}^{N}\log\left(1-\frac{1}{n^2}\right)=\log\prod_{n=2}^{N}\left(1-\frac{1}{N}\right)\left(1+\frac{1}{N}\right) = \log\frac{N+1}{2N}$$ so the series converges to $\color{red}{-\log 2}$.