How to evaluate the sum $\sum_{k=2}^{\infty}\log{(1-1/k^2)}$?

Question

How to evaluate the sum $$\sum\limits_{k=2}^{\infty}\log{(1-1/k^2)}\;?$$

Answer 1

Hint:

$$\log(1 - 1/k^2) = [\log(k+1) - \log(k)] - [\log(k) - \log(k-1)]$$

Answer 2

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 $$\prod_{k=2}^{\infty}(1-\frac{1}{k^2})=1/2$$ $$\sum_{k=2}^{\infty }\log(1-\frac{1}{k^2})=\color{red}{\log(1/2)}$$

Answer 3

$$\sum_{k=2}^{+\infty}\log\left(1-\frac{1}{k^2}\right)=\log\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)\tag{1}$$ 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: $$\sum_{k=2}^{+\infty}\log\left(1-\frac{1}{k^2}\right)=\log\frac{1}{2}=\color{red}{-\log 2}.\tag{3}$$