Matematické Fórum

Marian · 30. 05. 2016 17:42

Let L be an arbitrary positive integer. Find the closed evaluation of the infinite series

$\boldsymbol{\sum_{\substack{j=1\\j\neq L}}^{\infty}\frac{1}{ j^2-L^2}}.$

Bati · 31. 07. 2016 17:03

Hi ↑ Marian:,
it is an easy consequence of the cotangent expansion you mentioned in some other topic:
$\pi\cot{\pi x}=\frac1x+\sum_{n=1}^{\infty}\left(\frac1{n+x}-\frac1{n-x}\right)=\frac1x-2x\sum_{n=1}^{\infty}\frac1{n^2-x^2}$ , $x\in\mathbb{R}\setminus\mathbb{Z}$ .
From that we get
$\sum_{\substack{n=1\\n\neq L}}^{\infty}\frac{1}{n^2-x^2}=\frac{\frac1x-\pi\cot\pi x}{2x}-\frac1{L^2-x^2}$ for every $x\in\mathbb{R}\setminus\mathbb{Z}$ . Since the left hand side is continuous at $L$ , it remains to compute the limit of the right hand side. After several applications of the L'Hospital's rule, I obtained the result $\frac3{4L^2}$ .

Matematické Fórum

#1 30. 05. 2016 17:42

Infinite series with a gap

#2 31. 07. 2016 17:03

Re: Infinite series with a gap

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