Sum related to zeta function

Question

I was trying to evaluate the following sum: $$\sum_{k=0}^{\infty} \frac{1}{(3k+1)^3}$$

W|A gives a nice closed form but I have zero idea about the steps involved to evaluate the sum. How to approach such sums?

Following is the result given by W|A: $$\frac{13\zeta(3)}{27}+\frac{2\pi^3}{81\sqrt{3}}$$

Any help is appreciated. Thanks!

Answer 1

Notice that $$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{1}{27} \sum_{n=0}^{\infty} \frac{1}{(k+\frac{1}{3})^{3}} = - \frac{1}{54} \psi_{2}\left(\frac{1}{3} \right) $$

where $\psi_{2}(x)$ is the second derivative of the digamma function.

Differentiating the multiplication formula for the digamma function twice and letting $q=3$,

$$\psi_{2}(x) + \psi_{2} \left( x+ \frac{1}{3} \right) + \psi_{2} \left(x+ \frac{2}{3} \right) = 27 \psi_{2}(3x) .$$

Therefore, $$ \begin{align} \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) &= 27 \psi_{2}(1) - \psi_{2}(1) \\ &=27 \left( -2 \zeta(3) \right) + 2 \zeta(3) = -52 \zeta(3). \tag{1}\end{align} $$

And differentiating the reflection formula for the digamma function twice,

$$ \psi_{2} (x) - \psi_{2}(1-x) = - 2\pi^{3} \cot(\pi z) \csc^{2}(\pi z) .$$

Therefore, $$\psi_{2} \left(\frac{1}{3} \right) - \psi_{2} \left( \frac{2}{3}\right) = - \frac{8 \pi^{3}}{3 \sqrt{3}} . \tag{2}$$

Adding $(1)$ and $(2)$,

$$ \psi_{2} \left( \frac{1}{3}\right) = -26 \zeta(3) - \frac{4 \pi^{3}}{3 \sqrt{3}} .$$

So

$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{13 \zeta(3)}{27} + \frac{2 \pi^{3}}{81 \sqrt{3}} .$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{}$ \begin{align} \color{#66f}{\large\sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}^{3}}} &={1 \over 27}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1/3}^{3}} =\left. -\,{1 \over 27}\,\partiald{}{\mu}\sum_{k = 0}^{\infty} {1 \over \pars{k + \mu}\pars{k + 1/3}}\,\right\vert_{\,\mu\ =\ {1/3}} \\[3mm]&=-\,{1 \over 27}\,\partiald{}{\mu}\bracks{% \Psi\pars{\mu} - \Psi\pars{1/3} \over \mu - 1/3}_{\mu\ =\ {1/3}} =-\,{1 \over 54}\,\Psi''\pars{1 \over 3} \\[3mm]&=\color{#66f}{\large{1 \over 243}\bracks{2\root{3}\pi^{3} + 117\zeta\pars{3}}} \approx 1.0208 \end{align}

See a Hurwitz Zeta Function link.

Answer 3

You can also start from $$\sum_{k=0}^{m} \frac{1}{(3k+1)^3}=\frac{1}{54} \left(\psi ^{(2)}\left(m+\frac{4}{3}\right)-\psi ^{(2)}\left(\frac{1}{3}\right)\right)$$ which simplifies to $$\sum_{k=0}^{m} \frac{1}{(3k+1)^3}=\frac{1}{54} \left(\psi ^{(2)}\left(m+\frac{4}{3}\right)+26 \zeta (3)+\frac{4 \pi ^3}{3 \sqrt{3}}\right)$$and take the limit for an infinite value of $m$. This leads to the answer given by Felix Marin and by Wolfram Alpha.

In fact, there is a nice generalization for $$\sum_{k=0}^{\infty} \frac{1}{(ak+b)^c}=a^{-c} \zeta \left(c,\frac{b}{a}\right)$$ in which appears Hurwitz Zeta function.