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This question already has an answer here:

Find the value of $$\sum_{n=2}^{\infty}\log\left(1-\frac{1}{n^2}\right)$$

I tried expressing the sum like $\sum a_r-a_{r-1}$. $$\sum_{n=2}^{\infty}\log\left(1-\frac{1}{n^2}\right)=\sum_{n=2}^{\infty}\log\left[\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\right]=\sum_{n=2}^{\infty}\log\left(\frac{n-1}{n}\right)-\log\left(\frac{n}{n+1}\right)$$

I got stuck here. Is there any other simpler method?

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marked as duplicate by Guy Fsone, Rolf Hoyer, user99914, Misha Lavrov, Claude Leibovici Oct 31 '17 at 4:41

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    $\begingroup$ Would you know how to deal with $\sum_{n=2}^\infty \left( \log\left(n-1\right)-\log n - (\log n - \log(n+1) ) \right)$? $\endgroup$ – Clement C. Feb 24 '16 at 16:14
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Hint:

The last sum is telescopic.

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substitute the value of n from 2 and see all terms cancel out and some term will remain then let the n go to infinity and you will get an approx. answer.

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