Rylan Duncan

2022-01-23

Sum of series:
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4{n}^{2}-1}$

utgyrnr0

Expert

We want to compute it fastly and use the formula
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}-{x}^{2}}=\frac{1}{2{x}^{2}}-\frac{\pi \mathrm{cot}\pi x}{2x}$
where $x=\frac{1}{2}$ because
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{4{k}^{2}-1}=\frac{1}{4}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}-{\left(\frac{1}{2}\right)}^{2}}$

Amari Larsen

Expert

Hint: Partial Fraction decomposition:
$\frac{1}{4{n}^{2}-1}=\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$
You must then compute the closed form of
$\sum _{n=1}^{k}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$
Can you do that? Note that
$\sum _{n=1}^{k}\frac{1}{2n-1}=\frac{1}{1}+\frac{1}{3}+\dots +\frac{1}{2k-1}$
$=\frac{1}{2\cdot 0+1}+\frac{1}{2\cdot 1+1}+\dots +\frac{1}{2\left({k}_{1}\right)+1}$
$=\sum _{n=0}^{k-1}\frac{1}{2n+1}$
$=\sum _{n=1}^{k}\frac{1}{2n+1}+\frac{1}{2\cdot 0+1}-\frac{1}{2k+1}$

RizerMix

Expert

Hint: Work on ${S}_{n}=\sum _{k=1}^{n}\frac{1}{4{k}^{2}-1}$ and take its limit when $n\to \mathrm{\infty }$. Note that $\frac{1}{4{n}^{2}-1}=\frac{1}{2\left(2n-1\right)}-\frac{1}{2\left(2n+1\right)}$