Proving the convergence of $\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$ How to prove that the series
$$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$
is convergent? What about finding the sum?
My attempt:
$$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$
The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.

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