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$$\sum_{n=2}^{\infty}\ln\left(1-\frac{1}{n^2}\right) $$

I think it should be like that:

$$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right) $$ than I try to rewrite it: $$\sum_{n=2}^{k}\ln\left(\frac{n^2-1}{n^2}\right)=\ln(3)-\ln(4)+\ln(8)-\ln(9)...+\ln(k^2-1)-\ln(k^2) $$

I don't know what to do next. When I was doing other examples I had situation like this: $1+2-2+3-3+4-4$ etc. so I had 1st component-last component so it was a little bit easier.

2 Answers 2

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Observe that $$\ln \left(\frac{n^2-1}{n^2}\right)$$ $$=\ln \left(\frac{(n-1)(n+1)}{n^2}\right)$$ $$=\ln (n-1) +\ln (n+1) -2\ln n$$ $$=[\ln (n+1) -\ln n]-[\ln n -\ln (n-1)]$$

So $$\sum_{n=2}^\infty \ln \left(\frac{n^2-1}{n^2}\right)$$ $$=\sum_{n=2}^\infty[\ln (n+1) -\ln n]-\sum_{n=2}^\infty[\ln n -\ln (n-1)]$$ $$=\sum_{n=3}^\infty[\ln n -\ln (n-1)]-\sum_{n=2}^\infty[\ln n -\ln (n-1)]$$

Hope this helps you and you can solve it now.

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    so final answer is ln(1)-ln(n+1), limit of this is - $\infty$ so it is divergent with no sum – 2017-02-24
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    Do your sum carefully. It is a convergent sum. Answer is $-\ln 2$. – 2017-02-24
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    sorry, my bad. Thanks for your help. – 2017-02-24
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$$\sum_{r=2}^n\ln\dfrac{r^2-1}{r^2}=\ln\prod_{r=2}^n\dfrac{r^2-1}{r^2}$$

$$\prod_{r=2}^n\dfrac{r^2-1}{r^2}=\prod_{r=2}^n\dfrac{T_r}{T_{r+1}}=?$$ where $T_m=\dfrac{m-1}m$