Get the answer to "Proof of∑n=1∞1n3sinhπn2−sin⁡πn2coshπn2−cos⁡πn2=π3182"

ashes86047xhz

Answered question

2022-03-31

Proof of

\sum_{n = 1}^{\infty} \frac{1}{n^{3}} \frac{sinh π n \sqrt{2} - \sin π n \sqrt{2}}{cosh π n \sqrt{2} - \cos π n \sqrt{2}} = \frac{π^{3}}{18 \sqrt{2}}

Harry Gibson

Beginner2022-04-01Added 13 answers

It is a simple trigonometric exercise to show that

\frac{1}{n^{3}} \frac{sinh π n \sqrt{2} - \sin π n \sqrt{2}}{cosh π n \sqrt{2} - \cos π n \sqrt{2}} = R (\coth π n z_{0} + z_{0}^{2} \coth \frac{π n}{z_{0}})

(1)
with

z_{0} = e^{i \frac{π}{4}}

Recall the Ramanujan identity :

\sum_{n = 1}^{\infty} \frac{1}{n^{3}} (\coth π n x + x^{2} \coth \frac{π n}{x}) = \frac{π^{3}}{90 x} (x^{4} + 5 x^{2} + 1)

(2)
Combining (1) and (2), we immediately get

\sum_{n = 1}^{\infty} \frac{1}{n^{3}} (\frac{sinh π n \sqrt{2} - \sin π n \sqrt{2}}{cosh π n \sqrt{2} - \cos π n \sqrt{2}}) = \frac{π^{3}}{90} R \frac{z_{0}^{4} + 5 z_{0}^{2} + 1}{z_{0}}

= \frac{π^{3}}{18 \sqrt{2}}

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