Explained: "How to show that∑n=1∞1n2+3n+1=π55tan⁡π52"

klancimarsn

Answered question

2022-02-24

How to show that

\sum_{n = 1}^{\infty} \frac{1}{n^{2} + 3 n + 1} = \frac{π \sqrt{5}}{5} \tan \frac{π \sqrt{5}}{2}

poskakatiqzk

Beginner2022-02-25Added 8 answers

I think we can make some use of the residue theorem. Write

n^{2} + 3 n + 1 = {(n + \frac{3}{2})}^{2} - \frac{5}{4}

and the sum

\sum_{n = 1}^{\infty} \frac{1}{{(n + \frac{3}{2})}^{2} - \frac{5}{4}} = \frac{1}{2} \sum_{n = - \infty}^{\infty} \frac{1}{{(n + \frac{3}{2})}^{2} - \frac{5}{4}} - 1 + 1

(To get the doubly infinite sum, I had to add back the

n = 0

and

n = - 1

terms, which happen to sum to zero.)
The sum on the RHS may be attacked via the residue theorem, using the following:

\sum_{n = - \infty}^{\infty} f (n) = - π \sum_{k} R e s [f (z) \cot π z]

where

z_{k}

is a non-integer pole of f. The poles are at

z_{\pm} = - \frac{3}{2} \pm \frac{\sqrt{5}}{2}

The sum is then equal to

\frac{π}{2 \sqrt{5}} [\cot (\frac{3 π}{2} - \frac{\sqrt{5} π}{2}) - \cot (\frac{3 π}{2} + \frac{\sqrt{5} π}{2})] = \frac{π}{\sqrt{5}} \tan \frac{\sqrt{5} π}{2}

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