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I am not sure how to evaluate the infinite sum: $$\sum_{n=0}^\infty \frac{1}{(2n+1)^6}$$

Apparently, I can shift it to $$\sum_{n=1}^\infty \frac{1}{(2n-1)^6}$$ which is supposed to be a well known sum that is equal to $\frac{\pi^6}{960}$. However I can't find the proof for this. Aside from this method, I found that I can also use the fourier series, but I do not know how to do this. I would greatly appreciate your help. Thank you.

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    $\begingroup$ Do you know this trick? (look at highest score answer by user17762). $\endgroup$ Commented Sep 23, 2022 at 13:58
  • $\begingroup$ In general, for $k=1,2,...$ $$\sum_{n=0}^\infty(2n+1)^{-2k}=\sum_{n=1}^\infty\bigg((2n-1)^{-2k}+(2n)^{-2k}-(2n)^{-2k}\Big))$$ $$=\zeta(2k)-2^{-2k}\zeta(2k)=(1-2^{-2k})\zeta(2k)$$ $\endgroup$
    – Svyatoslav
    Commented Sep 23, 2022 at 14:02
  • $\begingroup$ Yes, or simply: for $m=2,3,\dots$$$\sum_{n=0}^\infty\frac1{(2n+1)^m}=\zeta(m)-\sum_{n=1}^\infty\frac1{(2n)^m}=\left(1-\frac1{2^m}\right)\zeta(m).$$ $\endgroup$ Commented Sep 23, 2022 at 14:12

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We know that $$\sum_{n=1}^\infty n^{-6}=\zeta(6)=\frac{\pi^6}{945}$$ This is absolutely convergent, so we can divide by $2^6$ to get the sum of even-$n$ terms: $$\sum_{n=1}^\infty(2n)^{-6}=\frac1{64}\zeta(6)$$ Subtracting this from the original gives the desired answer (on odd-$n$ terms): $$\sum_{n=0}^\infty(2n+1)^{-6}=\frac{63}{64}\zeta(6)=\frac{\pi^6}{960}$$

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