A sine series

Calculus (Integrals, Series)
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Tolaso J Kos
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A sine series

#1

Post by Tolaso J Kos »

Prove that:

$$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin{\sqrt{2} \pi n}} = -\frac{13 \pi^3}{360 \sqrt{2}}$$
Imagination is much more important than knowledge.
galactus
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Re: A sine series

#2

Post by galactus »

Yes, I see now. This is one of those tricky SOB's. :roll: :oops:

I used the wrong f(z) a while ago anyway.

I ran it through Maple for a check and it returns the correct result that you posted.


The residue at 0 is $$\frac{-\pi^{3}\sqrt{2}}{36}$$

$$f(z)=\frac{\pi cot(\pi z)csc(\pi \sqrt{2}z)}{z^{3}}$$

Let S be the sum in question, then note that:

$$S=Res(f(z), 0)+\frac{\sqrt{2}}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{4}}$$

$$\frac{-\pi^{3}\sqrt{2}}{36}+\frac{7\pi^{3}\sqrt{2}}{720}=\frac{-13\pi^{3}\sqrt{2}}{720}=\frac{-13\pi^{3}}{360\sqrt{2}}$$
Last edited by galactus on Sun Mar 06, 2016 8:41 pm, edited 4 times in total.
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Tolaso J Kos
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Re: A sine series

#3

Post by Tolaso J Kos »

Νο C,

the result I stated is correct. Mathematica and Wolfram on the other hand get it wrong.
Imagination is much more important than knowledge.
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