Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?

Question

Can help me to find $\sum_{n=1}^{\infty }\frac{1}{(4n-1)^3}$?

Answer 1

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".

Answer 2

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

$$\frac{(1-2^{-3})\zeta(3)-L(\chi, 3)}{2} = \frac{7}{16}\zeta(3) - \frac{\pi^3}{64}.$$