Compute $\sum_{n=1}^\infty n^{-6}$ using $\sum_{k=0}^\infty (2k+1)^{-6} = \frac{\pi^6}{960}$

Question

I have shown that $\sum_{k=0}^\infty (2k+1)^{-6} = \frac{\pi^6}{960}$, and from this I want to conclude that $\sum_{n=1}^\infty n^{-6} = \frac{\pi^6}{945}$, so I have been trying to show that

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

but I can't think of a simple way of doing so. Any help is welcome.

Answer 1

Let $S=\zeta(6)$ be the value of the sum you are trying to calculate. Notice that $$\begin{align} S=\sum_{n=1}^\infty \frac{1}{n^6} &=\sum_{n=0}^\infty \frac{1}{(2n+1)^6}+\sum_{n=0}^\infty \frac{1}{(2n)^6}\\ &=\sum_{n=0}^\infty \frac{1}{(2n+1)^6}+\frac{1}{2^6}\sum_{n=0}^\infty \frac{1}{n^6}\\ &=\frac{\pi^6}{960}+\frac{S}{2^6}\\ &=\frac{\pi^6}{960}+\frac{S}{64} \end{align}$$ Now you can easily solve the equation $$S=\frac{\pi^6}{960}+\frac{S}{64}$$ to find the value of your series.

Answer 2

More generally, if $S_m =\sum_{k=0}^\infty (2k+1)^{-m} $ and $T_m =\sum_{k=1}^\infty k^{-m} $, then

$\begin{array}\\ T_m &=\sum_{k=1}^\infty k^{-m}\\ &=\sum_{k=1}^\infty ((2k-1)^{-m}+(2k)^{-m})\\ &=\sum_{k=1}^\infty (2k-1)^{-m}+\sum_{k=1}^\infty(2k)^{-m}\\ &=\sum_{k=0}^\infty (2k+1)^{-m}+2^{-m}\sum_{k=1}^\infty k^{-m}\\ &=S_m+2^{-m}T_m\\ \text{so}\\ T_m &=\dfrac{S_m}{1-2^{-m}}\\ &=\dfrac{2^mS_m}{2^m-1}\\ \end{array} $

Yours is the case $m=6$.