sum of this series: $\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$

Note $\frac{1}{4n^2-1}=\frac{1}{(2n+1)(2n-1)}={\frac{1}{2}}\times\frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={\frac{1}{2}}\times[\frac{1}{2n-1}-\frac{1}{2n+1}]$ for $n\in\mathbb N.$

Let for $k\in\mathbb N,$ $S_k=\displaystyle\sum_{n=1}^{k}\frac{1}{4n^2-1}$ $\implies S_k={\frac{1}{2}}\displaystyle\sum_{n=1}^{k}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right].$ Thus for $k=1,2,...$

$S_1={\frac{1}{2}}\displaystyle\sum_{n=1}^{1}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}(1-\frac{1}{3})$

$S_2={\frac{1}{2}}\displaystyle\sum_{n=1}^{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})]=\frac{1}{2}(1-\frac{1}{5})$

$S_3={\frac{1}{2}}\displaystyle\sum_{n=1}^{3}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})]=\frac{1}{2}(1-\frac{1}{7})$

...

$S_k=\frac{1}{2}(1-\frac{1}{2k+1})$

$\implies\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\displaystyle\lim_{k\to\infty}S_k=\frac{1}{2}.$

Hint: Partial Fraction decomposition:$$\frac{1}{4n^2-1}=\frac{1}{(2n-1)(2n+1)}=\frac12[\frac{1}{2n-1}-\frac{1}{2n+1}]$$ You must then compute the closed form of $$\sum_{n=1}^k[\frac{1}{2n-1}-\frac{1}{2n+1}]$$ Can you do that? Note that $$\sum_{n=1}^k\frac{1}{2n-1}=\frac11+\frac13+...+\frac1{2k-1}=\frac1{2\cdot 0+1}+\frac1{2\cdot 1+1}+...+\frac1{2(k-1)+1}=\sum_{n=0}^{k-1}\frac{1}{2n+1}=\sum_{n=1}^{k}\frac1{2n+1}+\frac{1}{2\cdot 0+1}-\frac1{2k+1}$$

Or we want to compute it fastly and use the formula $$\sum_{k=1}^\infty \frac1{k^2-x^2}=\frac1{2x^2}-\frac{\pi\cot\,\pi x}{2x}$$ where $x=\frac{1}{2}$ because $$\sum_{k=1}^{\infty}\frac{1}{4k^2-1}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2-\left(\frac{1}{2}\right)^2}$$ Here you may find more information about this precious way.

Hint: Work on $S_n=\sum_{k=1}^n\frac{1}{4k^2-1}$ and take its limit when $n\to\infty$. Note that $$\frac{1}{4n^2-1}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$$