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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!

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So far, the answers have amounted to demonstrating that the Hurwitz zeta function is (almost) the same as the polygamma function (see Brad's comment below), and then appealing to knowledge of that. Can anyone provide a direct derivation? – Semiclassical Jul 17 '14 at 4:37
Also: judging from equations (11) and (12) of the Polygamma page at Mathworld, the result given above has a broad generalization. Can anyone prove result directly? – Semiclassical Jul 17 '14 at 4:43
On a related note, $~\displaystyle\sum_{n=-\infty}^\infty\frac1{n+a}=\pi\cdot\cot(\pi a).~$ Differentiating twice with regard to a, we have $~\displaystyle\sum_{n=-\infty}^\infty\frac1{(n+a)^3}=\pi^3\cdot\cot(\pi a)\cdot\csc^2(\pi a).~$ Letting $a=\dfrac13$ , we get $\dfrac{4\pi^3}{3\sqrt3}$ , which is double the value of your second term, save for a factor of $\dfrac1{3^3}=\dfrac1{27}$ – Lucian Jul 17 '14 at 10:21
up vote 7 down vote accepted

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 $(2)$,

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


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

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You may be interested in knowing that the general multiplication and reflection formulas are available on the polygamma wikipedia page. – Brad Jul 17 '14 at 5:14
@Brad Is that a new page? I don't think I've seen that page before. – Random Variable Jul 17 '14 at 5:19
Apparently it has been around since 2003 but its possible that it wasn't linked at the bottom of the digamma page for very long. – Brad Jul 17 '14 at 5:22
I liked this one. Thank you for the answer! :) – Pranav Arora Jul 17 '14 at 5:35
@PranavArora $$ \psi''(z)=-2\sum_{n=0}^{\infty} \frac{1}{(n+z)^3} \implies \sum_{n=0}^{\infty} \frac{1}{(n+z)^{3}} = - \frac{\psi''(z)}{2}$$ – Random Variable Jul 17 '14 at 6:12

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{}$ \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.

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Thank you sir for the response but unfortunately, I do not understand most of it. :( How do you get $$\sum_{k=0}^{\infty} \frac{1}{(k+\mu)(k+1/3)}=\frac{\psi(\mu)-1/3}{\mu-1/3}$$? Also, how do you find the value of $\psi''(1/3)$? – Pranav Arora Jul 17 '14 at 4:25
@PranavArora The sum is an identity. See identity $\bf\mbox{6.3.16}$ in this table. Unfortunately, I found the last result with a symbolic software ( WA ). Indeed, I would like to derive the last formula. – Felix Marin Jul 17 '14 at 4:35
@PranavArora WA result – Felix Marin Jul 17 '14 at 4:36
For reference, $$\psi^{(m)}(z)= (-1)^{m+1} m! \zeta (m+1,z)$$ – Brad Jul 17 '14 at 4:38
+1 and nice formatting... – draks ... Jul 17 '14 at 5:14

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.

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