Rylan Duncan

Answered

2022-01-23

Sum of series:

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

Answer & Explanation

utgyrnr0

Expert

2022-01-24Added 11 answers

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

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{4{k}^{2}-1}=\frac{1}{4}\sum _{k=1}^{\mathrm{\infty}}\frac{1}{{k}^{2}-{\left(\frac{1}{2}\right)}^{2}}$

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where

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Amari Larsen

Expert

2022-01-25Added 10 answers

Hint: Partial Fraction decomposition:

$\frac{1}{4{n}^{2}-1}=\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$

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

$\sum _{n=1}^{k}\frac{1}{2n-1}=\frac{1}{1}+\frac{1}{3}+\dots +\frac{1}{2k-1}$

$=\frac{1}{2\cdot 0+1}+\frac{1}{2\cdot 1+1}+\dots +\frac{1}{2\left({k}_{1}\right)+1}$

$=\sum _{n=0}^{k-1}\frac{1}{2n+1}$

$=\sum _{n=1}^{k}\frac{1}{2n+1}+\frac{1}{2\cdot 0+1}-\frac{1}{2k+1}$

You must then compute the closed form of

Can you do that? Note that

RizerMix

Expert

2022-01-27Added 437 answers

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
$\frac{1}{4{n}^{2}-1}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$

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