Find sum of series I need to find the sum of the following series:
$$\sum_{n=2}^\infty \ln\left(1-\frac 1{n^2}\right)$$
How to proceed with this?
 A: Since:
$$\prod_{n=2}^{N}\left(1-\frac{1}{n^2}\right)=\prod_{n=2}^{N}\frac{n-1}{n}\cdot\prod_{n=2}^{N}\frac{n+1}{n}=\frac{1}{N}\cdot\frac{N+1}{2}$$
we have:
$$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right) = \color{red}{-\log 2}.$$
It is also interesting to prove that:
$$\sum_{n\geq 2}\log\left(1+\frac{1}{n^2}\right) = \log\left(\frac{\sinh \pi}{2\pi}\right).$$
A: Hint: sum of logs = log of product.  This is a very famous infinite product.
A: Another way to compute the product.
$$\sin(z) = z\prod_{n=1}^\infty \left(1-\frac{z^2}{\pi^2n^2}\right)$$
$$\frac{\sin(z)}{z(1-\frac{z^2}{\pi^2})} = \prod_{\color{red}{n=2}}^\infty \left(1-\frac{z^2}{\pi^2 n^2}\right)$$
$$\lim_{z\rightarrow \pi}\frac{\sin(z)}{z(1-\frac{z}{\pi})(1+\frac{z}{\pi})} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi}\frac{\color{blue}{-(z-\pi)}+ \color{green}{(z-\pi)^3/6+\cdots}}{(1-\frac{z}{\pi})} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi}\frac{\color{blue}{-(z-\pi)}+\color{green}{(z-\pi)^3/6+\cdots}}{-\frac{1}{\pi}(z-\pi)} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi} \color{blue}{ \frac{ -(z-\pi) }{-\frac{1}{\pi} (z-\pi)}} + \color{green}{\frac{(z-\pi)^3/6+\cdots }{-\frac{1}{\pi} (z-\pi)}}  = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi} \color{blue}{ \pi} +\color{green}{ \pi(z-\pi)^2/6+\cdots } = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi} (\color{blue}{\pi} + \color{green}{0} )= \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi} \color{blue}{\pi}= \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
So $$ \sum_{n=2}^\infty \log(1-\frac{1}{n^2}) = -\log(2)$$
