# How to find the sum of this infinite series

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.

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

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): $$\sum_{n=0}^\infty(2n+1)^{-6}=\frac{63}{64}\zeta(6)=\frac{\pi^6}{960}$$