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

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?

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

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