# How can we evaluate $\sum_{k\geq 0} \frac{1}{(2k+1)^3}$?

I have been looking to evaluate $$\mathcal{A} = \sum_{k=0}^\infty \frac{1}{(2k+1)^3}.$$ We can represent our sum in terms of the Hurwitz zeta function; namely, $$\mathcal{A} = \zeta\left(\frac{1}{2}, 3\right) = \frac{1}{8}\sum_{k=0}^\infty \frac{1}{\left(k+\frac{1}{2}\right)^2}.$$ And from here , we know that
$$\frac{\psi^{\left(-1/2\right)}(3)}{\sqrt{\pi i}} = \zeta\left(\frac{1}{2}, 3\right)$$ which I have no idea how to compute. I am sure there is a less cumbersome way to evaluate this sum. The answer to the sum is $$\frac{7}{8}\zeta\left(3\right)$$ which seems like it would be a standard computation. Any help would be greatly appreciated.

$$\zeta(3)=\sum_{n=1}^\infty\frac1{(2n)^3}+\sum_{n=0}^\infty\frac1{(2n+1)^3}=\frac18\sum_{n=1}^\infty\frac1{n^3}+\sum_{n=0}^\infty\frac1{(2n+1)^3}$$
Note that for $$s>1$$ $$\sum_{k\ge0}\frac1{(2k+1)^s}=\sum_{k\ge1}\frac1{k^s}-\sum_{k\ge1}\frac1{(2k)^s}=\sum_{k\ge1}\frac1{k^s}-\frac1{2^s}\sum_{k\ge1}\frac1{k^s}=\left(1-\frac1{2^s}\right)\zeta(3)$$ Set $$s=3$$ to obtain your result.
$$\sum_{k=1}^{\infty}\dfrac{1}{k^3} - \sum_{k =1}^{\infty}\dfrac{1}{(2k)^3}= \left(1 - \dfrac{1}{8}\right)\sum_{k =1}^{\infty}\dfrac{1}{k^3}= \dfrac{7}{8}\zeta(3)$$