Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$ I want to prove that
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
My approach is the most logical one, rewrite as follows:
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $
But the remaining series are both divergent so I get to a sort of "$ \infty - \infty = 1 $".
 A: $$\lim_{n\to\infty}\left(\sum_{r=1}^{n} \frac{1}{2r-1} - \sum_{r=1}^{n} \frac{1}{2r+1}\right) $$
$\require{cancel}$
$$=\lim_{n\to\infty}\left(\frac11-
 \cancel{\frac13}+\cancel{\frac13}-\cancel{\frac15}+\cancel{\frac15}-\cdots+\cancel{\frac{1}{2n-1}} -  \frac{1}{2n+1}\right)$$
$$=\lim_{n\to\infty}\frac{2n}{2n+1}$$
I'll leave it to you to take the limit
This is known as the method of differences
A: Rewriting the term,
$\color{blue}{\left(\sum_{n=1}^{\infty} \frac{2}{4 n^{2} - 1}\right)}=\color{black}{\left(\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)\right)}$
This is the telescoping series:
$\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)=\left(1-\color{green}{\frac{1}{3}}\right)+\left(\color{green}{\frac{1}{3}}-\color{red}{\frac{1}{5}}\right)+\left(\color{green}{\frac{1}{5}}-\color{red}{\frac{1}{7}}\right)+\left(\color{blue}{\frac{1}{7}}-\color{red}{\frac{1}{9}}\right)+...=1$
