Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$? Can help me to find  $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?
 A: 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
$$\frac{(1-2^{-3})\zeta(3)-L(\chi, 3)}{2} = \frac{7}{16}\zeta(3) - \frac{\pi^3}{64}.$$
A: To compute this series let's rewrite it first as a polygamma function :
\begin{align}
\tag{1}S&=\sum_{k=0}^{\infty }\frac{1}{(4k+3)^3}\\
S&=\frac 1{4^3}\sum_{k=0}^{\infty }\frac{1}{\left(k+\frac 34\right)^3}\\
\tag{2}S&=-\frac 1{2!\,4^3}\psi^{(2)}\left(\frac 34\right)\\
\end{align}
Since (from the previous link) $$\tag{3} \psi^{(2)}(z)=-2!\sum_{k=0}^{\infty }\frac{1}{\left(k+z\right)^3}$$  
Computing the second derivative of the logarithm derivative of Euler's reflection formula for $\Gamma$ we get (since $\psi(z):=(\ln\,\Gamma(z))'\,$ and $\,(\ln\,\sin(\pi z))'=\pi\,\cot(\pi\,z)$ ) following reflection relation :
$$\tag{4}\psi^{(2)}(1-z)-\psi^{(2)}(z)=\pi\frac {d^2}{dz^2} \cot(\pi\,z)$$
that is for $z=\frac 14$ :
$$\tag{5}\psi^{(2)}\left(\frac 34\right)-\psi^{(2)}\left(\frac 14\right)=\lim_{z\to 1/4}\left[2\pi^3\cot(\pi\,z)(\cot(\pi\,z)^2+1)\right]=4\,\pi^3$$
But from $(3)$ we have too :
\begin{align}
\tag{6}\psi^{(2)}(z)+\psi^{(2)}\left(z+\frac 12\right)&=-2\left[\sum_{k=0}^{\infty }\frac{1}{\left(k+z\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(k+z+1/2\right)^3}\right]\\
\psi^{(2)}\left(\frac 14\right)+\psi^{(2)}\left(\frac 34\right)&=-2\left[\sum_{k=0}^{\infty }\frac{1}{\left(k+1/4\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(k+3/4\right)^3}\right]\\
&=-2\cdot 4^3\left[\sum_{k=0}^{\infty }\frac{1}{\left(4k+1\right)^3}+\sum_{k=0}^{\infty }\frac{1}{\left(4k+3\right)^3}\right]\\
&=-2\cdot 4^3\sum_{n=1}^{\infty }\frac{1}{\left(2n-1\right)^3}\\
&=-2\cdot 4^3\left[\sum_{n=1}^{\infty}\left(\frac{1}{\left(2n-1\right)^3}+\frac{1}{\left(2n\right)^3}\right)-\sum_{n=1}^{\infty}\frac{1}{\left(2n\right)^3}\right]\\
&=-2\cdot 4^3\left[\zeta(3)-\frac {\zeta(3)}8\right]\\
\tag{7}&=-2\cdot 4^3\frac 78\zeta(3)\\
\end{align}
Adding $(5)$ and $(7)$ we obtain $\,2\,\psi^{(2)}\left(\frac 34\right)\,$ at the left so that $(2)$ becomes :
$$S=-\frac 1{4\cdot 4^3}\cdot 2\,\psi^{(2)}\left(\frac 34\right)=-\frac 4{4\cdot 4^3}\pi^3+\frac{2\cdot 4^3}{4\cdot 4^3}\frac 78\zeta(3)$$
or simply
$$\tag{8}\boxed{\displaystyle S=\frac 7{16}\zeta(3)-\frac{\pi^3}{64}}$$
For a more general proof see Kölbig's $1996$ paper "The polygamma function $\psi^{(k)}(x)$ for $x=\frac 14$ and $x=\frac 34$".
For other rational arguments see "The polygamma function and the derivatives of the cotangent function for rational arguments".
