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Proving $$\sum_{n=1,3,5..}^{\infty }\frac{\sin^r(n\pi/3)}{n^2}=\frac{3^{0.5r-2}}{2^r}\pi^2$$ if the $r$ an even integer number greater than 0

I don't have the enough experience to prove formulas like the above, so I need some helps to prove it. Thanks

1 Answers1

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This is just: $$\sum_{n\geq 0}\left(\frac{1}{(6n+1)^2}+\frac{1}{(6n+5)^2}\right) =\frac{\pi^2}{9}\tag{1} $$ written in a very ugly way. To prove the last identity, notice that the LHS equals: $$ -\sum_{n\geq 0}\int_{0}^{1}\left(x^{6n}+x^{6n+4}\right)\log x\,dx = -\int_{0}^{1}\frac{1+x^4}{1-x^6}\,\log x\,dx\tag{2}$$ that can be computed with many different techniques.

Jack D'Aurizio
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  • Is your answer for special case or general? –  Feb 20 '15 at 21:57
  • @Reery: it applies just here since $\sin((2m+1)\pi/3)$ behaves like $\frac{\sqrt{3}}{2},0,-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},0,-\frac{\sqrt{3}}{2},\ldots$ but line $(2)$ provides a general technique for writing such sums into an "integral format". – Jack D'Aurizio Feb 20 '15 at 22:06
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    Like you said, these can be evaluated with many different techniques. One of them is the $\pi\cot(\pi z)$ multiplier technique which is well known and of which there is an example at this MSE link. – Marko Riedel Feb 20 '15 at 22:26