How to evaluate the sum $\sum_{k=2}^{\infty}\log{(1-1/k^2)}$? How to evaluate the sum $$\sum\limits_{k=2}^{\infty}\log{(1-1/k^2)}\;?$$
 A: we know that 
$$\sum \log(a)=\log\prod (a)$$
use $$\sin(x)=x\prod_{k=1}^{\infty}(1-\frac{x^2}{(k*\pi)^2})$$
$$\sin(x\pi)=x\pi\prod_{k=1}^{\infty}(1-\frac{x^2}{k^2})$$
$$\frac{\sin(x\pi)}{x}=\pi\prod_{k=1}^{\infty}(1-\frac{x^2}{k^2})$$
$$\frac{\sin(x\pi)}{x(1-x^2)}=\pi\prod_{k=2}^{\infty}(1-\frac{x^2}{k^2})$$
take the limit ( L'Hôpital's rule ) of the L.H.S at $x=1$ to get
$$\lim_{x\rightarrow 1}\frac{\sin(x\pi)}{x(1-x^2)}=\pi/2$$ 
hence
$$\prod_{k=2}^{\infty}(1-\frac{1}{k^2})=1/2$$
$$\sum_{k=2}^{\infty }\log(1-\frac{1}{k^2})=\color{red}{\log(1/2)}$$
A: $$\sum_{k=2}^{+\infty}\log\left(1-\frac{1}{k^2}\right)=\log\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)\tag{1}$$
but the product in the RHS of $(1)$ is telescopic:
$$\prod_{k=2}^{N}\left(1-\frac{1}{k^2}\right) = \prod_{k=2}^{N}\frac{k-1}{k}\prod_{k=2}^{N}\frac{k+1}{k}=\frac{N+1}{2N}\tag{2} $$
hence:
$$\sum_{k=2}^{+\infty}\log\left(1-\frac{1}{k^2}\right)=\log\frac{1}{2}=\color{red}{-\log 2}.\tag{3}$$
A: Hint:
$$\log(1 - 1/k^2) = [\log(k+1) - \log(k)] - [\log(k) - \log(k-1)]$$ 
