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 $-\log (2)$.
Can you please help me?
2026-04-01 12:31:01.1775046661
Calculate $\sum_{n\ge2}\log\left(1-\frac1{n^2}\right)$
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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.