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

Rylan Duncan

Rylan Duncan

Answered question

2022-01-23

Sum of series:
n=114n21

Answer & Explanation

utgyrnr0

utgyrnr0

Beginner2022-01-24Added 11 answers

We want to compute it fastly and use the formula
k=11k2x2=12x2πcotπx2x
where x=12 because
k=114k21=14k=11k2(12)2
Here you may find more information about this precious way.
Amari Larsen

Amari Larsen

Beginner2022-01-25Added 10 answers

Hint: Partial Fraction decomposition:
14n21=1(2n1)(2n+1)=12[12n112n+1]
You must then compute the closed form of
n=1k[12n112n+1]
Can you do that? Note that
n=1k12n1=11+13++12k1
=120+1+121+1++12(k1)+1
=n=0k112n+1
=n=1k12n+1+120+112k+1
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Hint: Work on Sn=k=1n14k21 and take its limit when n. Note that 14n21=12(2n1)12(2n+1)

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