Step by step solution to "How to get the sum of the series∑n=1∞1(4n2−1)2..."

Keri Molloy

Answered question

2022-02-24

How to get the sum of the series

\sum_{n = 1}^{\infty} \frac{1}{{(4 n^{2} - 1)}^{2}}

benehmenshgf

Beginner2022-02-25Added 7 answers

\sum_{n = 1}^{\infty} \frac{1}{{(4 n^{2} - 1)}^{2}} = \frac{1}{16} \frac{d}{d a} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - a^{2}} {|_{{a = \frac{1}{2}}} = \frac{1}{16} \frac{d}{d a} \sum_{n = 0}^{\infty} (\frac{1}{n + 1 - a} - \frac{1}{n + 1 + a}) \frac{1}{2 a} |}_{a = \frac{1}{2}}

= \frac{1}{32} \frac{d}{d a} (\frac{H_{a} - H_{- a}}{a}) ∣_{a = \frac{1}{2}}

= \frac{1}{32} \frac{d}{d a} (\frac{H_{a - 1} + \frac{1}{a} - H_{- a}}{a}) ∣_{a = \frac{1}{2}}

= \frac{1}{32} \frac{d}{d a} (\frac{1}{a^{2}} - \frac{π \cot (π a)}{a}) ∣_{a = \frac{1}{2}}

= \frac{1}{32} (- \frac{2}{a^{3}} + \frac{π \cot (π a)}{a^{2}} + \frac{π^{2} \csc^{2} (π a)}{a}) ∣_{a = \frac{1}{2}} = \frac{π^{2} - 8}{16}

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