Evaluating/Simplifying a Series involving Zeta Functions

Question

I was trying to learn how to evaluate this expression: $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}$$ And I learned that it can be easily simplified to: $$\frac{1}{64}(\zeta(3,\frac{1}{4})-\zeta(3,\frac{3}{4}))$$ using the Hurwitz Zeta function. From here, I didn't know how to further simplify it to get an answer of: $$\frac{\pi^{3}}{32}$$ On the Mathworld Wolfram page I had seen a simplification like this: $$\zeta(s,\frac{1}{2})=2^{s}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}=2^{s}(\zeta(s)-\sum_{k=1}^{\infty}\frac{1}{(2k)^{s}})=(2^{s}-1)\zeta(s)$$ But I had no idea how they even did this, or if this is somehow the way that can help to give the simplification to get the answer above.

Edit: I have realized that this is a Dirichlet Beta Function, but now I am wondering if still the original Hurwitz Zeta Function can be simplified without the use of definitions of the Beta Function. (As in, a connection to what I have written above with that simplification)

Answer 1

This is a Dirichlet beta function and your result may be obtained using this answer.

We may rewrite $\zeta\left(3, \pm \frac {1}4\right)$ using the polygamma function :

$$\tag{1}\psi^{(m)}(z) = (-1)^{m+1} m! \,\zeta(m+1,z)$$

You want $\displaystyle S:=\frac{1}{64}\left(\zeta\left(3,\frac{1}{4}\right)-\zeta\left(3,\frac{3}{4}\right)\right)=\frac{1}{64}\left(-\frac 12\psi^{(2)}\left( \frac {1}4\right)+\frac 12\psi^{(2)}\left( \frac {3}4\right)\right)$

These polygamma values are well known (from Kolbig's paper ) :

\begin{align} \tag{2}\psi^{(2n)}\left(\frac 14\right) &= -2^{2n-1}\left(\pi^{2n+1}|E_{2n}|+2(2n)!(2^{2n+1}-1)\zeta(2n+1)\right)\\ \psi^{(2)}\left(\frac 14\right)&=-2\left(\pi^{3}+2(2)!(2^{3}-1)\zeta(3)\right)\\ \tag{3}\psi^{(2n)}\left(\frac 34\right) &= 2^{2n-1}\left(\pi^{2n+1}|E_{2n}|-2(2n)!(2^{2n+1}-1)\zeta(2n+1)\right)\\ \psi^{(2)}\left(\frac 34\right)&=2\left(\pi^{3}-2(2)!(2^{3}-1)\zeta(3)\right)\\ \end{align}

Subracting the first to the second and dividing by $64$ gives indeed $\displaystyle\frac {\pi^3}{32}$.

We may get the result more directly (cf Kolbig) starting with this polygamma expression : $$\tag{4}\psi^{(n)}(z) = (-1)^{n+1}\, n! \sum_{k=0}^\infty \frac{1}{(z+k)^{n+1}}$$ using the polygamma reflection formula $6.4.7$ (itself derived from the $\Gamma$ function reflection formula $6.1.17$) : $$\tag{5} \psi^{(n)}\left(1-z\right)+(-1)^{n+1}\psi^{(n)}\left(z\right)=(-1)^n\pi \frac {d^n}{dz^n} \cot(\pi z)$$

So that $$\tag{6}\psi^{(2)}\left( \frac {3}4\right)-\psi^{(2)}\left( \frac {1}4\right)=\pi \left.\frac {d^2}{dz^2} \cot(\pi z)\right|_{z=1/4}$$ $$-$$ For a direct derivation use (proofs here ) : $$\tag{7} \pi \cot(\pi z)=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}=\frac 1z+\sum_{n=1}^\infty \frac 1{z-n}+\frac 1{z+n}$$ and compute the second derivative at $\displaystyle z=\frac 14$.