TITLE: sum of this series: $\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$
QUESTION [14 upvotes]: I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$
my steps: 
$$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum$$
I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..

REPLY [3 votes]: This is an easy problem by using Fourier's serie of $|\sin(x)|$. So,
$|\sin(x)|=\dfrac2\pi-\dfrac4\pi\sum_{n=1}^{\infty}\dfrac{\cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:
$0=\dfrac2\pi-\dfrac4\pi\sum_{n=1}^{\infty}\dfrac{1}{4n^2-1} $.
So,
$\sum_{n=1}^{\infty}\dfrac{1}{4n^2-1}=\dfrac12$