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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?

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  • $\begingroup$ 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 $N\geq 2$. $\endgroup$ Oct 13, 2018 at 16:07

1 Answer 1

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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.

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  • $\begingroup$ Okay. I solved it. I don't believe I did not remember n^2-1=(n+1)(n-1). That was my mistake because the rest was right. $\endgroup$ Oct 13, 2018 at 16:02
  • $\begingroup$ @AnaHelenaVieira It happens :-) $\endgroup$ Oct 13, 2018 at 16:05

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