TITLE: Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?
QUESTION [8 upvotes]: Can help me to find  $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?

REPLY [7 votes]: Here is a different approach. Let $\chi: (\mathbb Z/4\mathbb Z)^\times \to \{\pm 1\}$ be the Dirichlet character $a \mapsto (-1)^{a-1/2}$. 
The $L$-function $L(\chi, s) = \sum_{n\geq 1} \chi(n) n^{-s}$ satisfies the functional equation 
$$\Lambda(\chi, s) = \Lambda(\chi, 1-s)$$
where $$\Lambda(\chi, s) = (4/\pi)^{s/2} \Gamma\left(\frac{1+s}{2}\right) L(\chi, s).$$
For $n\geq 1$, we have
$$L(\chi, 1-n) = - \frac{B_{n, \chi}}{n}$$
where $${B_{n, \chi}} = {4^{n-1}}(B_n(1/4)-B_n(3/4)),$$
$B_n(X)$ the Bernoulli polynomial. Since $B_3(X) = X^3 - \frac32 x^2 + \frac12 x,$ we find that
$$L(\chi, -2) = -1/2.$$
Since $\Gamma(-1/2)= -  2 \sqrt \pi$, we find that
$$\Lambda(\chi, -2) = \frac{\pi^{3/2}}{4} = \Lambda(\chi, 3) = (4/\pi)^{3/2}L(\chi, 3)$$
hence 
$$L(\chi, 3) = \frac{\pi^3}{32}.$$
Now your number is
$$\frac{(1-2^{-3})\zeta(3)-L(\chi, 3)}{2} = \frac{7}{16}\zeta(3) - \frac{\pi^3}{64}.$$