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