# Infinite Series $‎\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$

‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎, $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ ‎Since if we set $C:=\sum_{m=2}^{\infty}|a_m|<\infty$‎, ‎then‎ $$\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m\frac{1}{n^m}|\leq\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}\leq C\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty‎$$ and ‎by Cauchy's double series theorem‎, ‎we can switch the order of summation‎: $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ This shows that $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}=\sum_{n=1}^{\infty}\frac{1}{kn(kn-1)}$.

My Questions:

1) It's obvious that $\sum_{n=1}^{\infty}\frac{1}{2n(2n-1)}=\log(2)$, but how can I evaluate $\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}$?

2) Is there another method to evaluate $\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$?

• Generally speaking, $$\sum_{n=1}^\infty\frac1{(n+a)(n+b)}=\frac{H_a-H_b}{a-b}$$ where $$H_n=\int_0^1\frac{1-x^n}{1-x}dx$$ More information available here. In your case, $a=0$ and $b=-\frac1k$, where k is $2$, $3$, etc. – Lucian Nov 13 '13 at 12:13

## 6 Answers

Extended Harmonic Numbers

Normally, we think of Harmonic Numbers as $$H_n=\sum_{k=1}^n\frac1k\tag{1}$$ However, an alternate definition is often useful: $$H_n=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+n}\right)\tag{2}$$ For integer $n\ge1$, it is not too difficult to see that the two definitions agree. However, $(2)$ is easily extendible to all $n\in\mathbb{R}$ (actually, to all $n\in\mathbb{C}$). We can say some things about $H_n$ for some $n\in\mathbb{Q}\setminus\mathbb{Z}$.

Note that for $m,n\in\mathbb{Z}$, \begin{align} H_{mn}-H_n &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+mn}\right)-H_n\\ &=\sum_{k=1}^\infty\sum_{j=0}^{m-1}\left(\frac1{km-j}-\frac1{km-j+mn}\right)-H_n\\ &=\frac1m\sum_{j=0}^{m-1}\sum_{k=1}^\infty\left(\left(\frac1k-\frac1{k-j/m+n}\right)-\left(\frac1k-\frac1{k-j/m}\right)\right)-H_n\\ &=\frac1m\sum_{j=0}^{m-1}\left(\left(H_{n-j/m}-H_n\right)-H_{-j/m}\right)\tag{3} \end{align} Since $H_0=0$ and $H_n=\log(n)+\gamma+O\left(\frac1n\right)$, where $\gamma$ is the Euler-Mascheroni Constant, if we let $n\to\infty$ in $(3)$, we get $$\sum_{j=1}^{m-1}H_{-j/m}=-m\log(m)\tag{4}$$ Using identity $(7)$ from this answer, \begin{align} \pi\cot(\pi z) &=\sum_{k\in\mathbb{Z}}\frac1{k+z}\\ &=\sum_{k=1}^\infty\left(\frac1{k-1+z}-\frac1{k-z}\right)\\ &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k-z}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z-1}\right)\\ &=H_{-z}-H_{z-1}\tag{5} \end{align} which implies $$H_{-j/m}-H_{-(m-j)/m}=\pi\cot\left(\frac{\pi j}{m}\right)\tag{6}$$

Using $(4)$ and $(6)$ for $m=3$ yields $$H_{-1/3}+H_{-2/3}=-3\log(3)\tag{7}$$ and $$H_{-1/3}-H_{-2/3}=\pi\cot\left(\frac\pi3\right)\tag{8}$$ Averaging $(7)$ and $(8)$ yields $$H_{-1/3}=-\frac32\log(3)+\frac\pi{2\sqrt3}\tag{9}$$ Finally, \begin{align} \sum_{n=1}^\infty\frac1{3n(3n-1)} &=-\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/3}\right)\\ &=-\frac13H_{-1/3}\\[6pt] &=\frac12\log(3)-\frac\pi{6\sqrt3}\tag{10} \end{align}

Values for Future Reference

Using $(4)$ and $(6)$, we can also compute \begin{align} H_{-1/4}&=\pi/2-3\log(2)\\ H_{-1/2}&=-2\log(2)\\ H_{-3/4}&=-\pi/2-3\log(2) \end{align}\tag{11}

• Creative and very nice! And easy – dust05 Sep 26 '13 at 17:14
• This deserves to be published. An MAA magazine seems appropriate. – marty cohen May 3 '15 at 16:34
• $H_{p/q}$ for $0\lt p/q\lt1$ can be computed using $(7)$ from this answer. – robjohn May 2 '17 at 16:12
• I don't recall how I found this answer, but after some searching, I believe this article / lecture notes by B Candelpergher shows that your $H_z$ is (or is related to) one of Ramanujan's summation methods. Chapter 1 definition 3 on page 22, and also the remark on page 24 hal.univ-cotedazur.fr/hal-01150208v2/document – Calvin Khor Sep 22 '19 at 6:10

Answer to Second Question

We remark that

$$\Gamma(z)\zeta(z)=\int_0^\infty \frac{u^{z-1}}{e^u-1}du\tag{1}$$ $$\sum_{n=2}^\infty \frac{t^n}{(n-1)!}=(e^t-1)t \tag{2}$$ $$\psi_0(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx \tag{3}$$ where $\psi_0(s)$ is Digamma Function and $\gamma$ is the Euler's Constant.

Then

\begin{align*} \sum_{n=2}^\infty \frac{\zeta(n)}{k^n} &= \sum_{n=2}^\infty \frac{1}{k^n \Gamma(n)}\int_0^\infty \frac{u^{n-1}}{e^u-1}du\\ &=\int_0^\infty\frac{1}{u(e^u-1)}\left( \sum_{n=2}^\infty \frac{1}{(n-1)!}\left(\frac{u}{k}\right)^n\right)du \\ &= \frac{1}{k}\int_0^\infty \frac{e^{\frac{u}{k}}-1}{e^u-1}du \tag{4} \end{align*}

Substituting $t=e^{-u}$, we get

\begin{align*} \sum_{n=2}^\infty \frac{\zeta(n)}{k^n}&=-\frac{1}{k}\int_0^1 \frac{1-t^{-1/k}}{1-t}dt \\ &= \frac{-1}{k}\left\{ \gamma+\psi_0 \left(1-\frac{1}{k} \right)\right\} \tag{5} \end{align*}

If $k(\geq 2)$ is an integer, equation $(5)$ can be further simplified using Gauss' Digamma Theorem.

$$\sum_{n=2}^\infty \frac{\zeta(n)}{k^n}=-\frac{\pi}{2k}\cot \left( \frac{\pi}{k}\right)+\frac{\log k}{k}-\frac{1}{k}\sum_{m=1}^{k-1}\cos \left(\frac{2\pi m}{k} \right) \log \left( 2\sin \frac{\pi m}{2}\right)$$

Answering you first question.
Don't know whether it's simpler but still (since you'll have to deal with hypergeometric functions). $$\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}=\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{3n-1}\int_0^1x^{n-1} \mathrm dx=\frac{1}{3}\int_0^1\sum_{n=1}^{\infty}\frac{x^{n-1}}{3n-1}\mathrm dx$$ $$\sum_{n=1}^{\infty}\frac{x^{n-1}}{3n-1}=\frac{1}{2} \, _2F_1\left(\frac{2}{3},1;\frac{5}{3};x\right)$$ And $$\frac{1}{6} \int_0^1 \, _2F_1\left(\frac{2}{3},1;\frac{5}{3};x\right) \, \mathrm dx=\frac{1}{6} \left(3\log(3) -\frac{\pi }{\sqrt{3}}\right)$$ So $$\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}=\frac{1}{6} \left(3\log(3) -\frac{\pi }{\sqrt{3}}\right)$$

This is an aswer for every $k$, using your method. For $k=2$ and $3$ it fits with other answers, so it's probably correct :) It is completely elementary, in the sense that it uses just the Taylor expansion of $\log(1-x)$ and the fact that the sum $\sum_{\alpha^k=1}\alpha^n$ ($\alpha$ runs over the $k$-th roots of $1$) is $k$ if $k$ divides $n$ and $0$ otherwise.

There are some $\log$'s of complex numbers. Those numbers have always non-negative real part, for the Arg we take the angle between $-\pi/2$ and $\pi/2$, so that it fits with the power series for $\log(1-x)$.

$$\sum_n x^{kn}/kn=-\log(1-x^k)/k=-\frac{1}{k}\sum_{\alpha^k=1}\log(1-\alpha x)$$ $$\sum_{\alpha^k=1}\sum_{m=1}^\infty\alpha(\alpha x)^m/m=k\sum_{n=1}^{\infty}x^{kn-1}/(kn-1)$$ but also $$\sum_{\alpha^k=1}\sum_{m=1}^\infty\alpha(\alpha x)^m/m=-\sum_{\alpha^k=1}\alpha\log(1-\alpha x).$$ We thus have $$\sum_n x^{kn}/(kn(kn-1))=\sum_n x^{kn}(\frac{1}{kn-1}-\frac{1}{kn})=$$ $$=\frac{1}{k}\sum_{\alpha^k=1}(1-x\alpha)\log(1-\alpha x).$$ We have to take the limit $x\to 1$. The $\alpha=1$ term disappears, so we get $$\sum_n\frac{1}{kn(kn-1)}=\frac{1}{k}\sum_{\alpha^k=1,\alpha\neq1}(1-\alpha)\log(1-\alpha)=$$ ($\alpha=\exp(2\pi i m/k)$) $$=\frac{1}{k}\sum_{m=1}^{k-1}((1-\cos\frac{2\pi m}{k})-i\sin\frac{2\pi m}{k})(\log(2\sin\frac{\pi m}{k})+\pi i(m/k-1/2))=$$ $$=\frac{1}{k}\sum_{m=1}^{k-1}(1-\cos\frac{2\pi m}{k})\log(2\sin\frac{\pi m}{k})+\pi(m/k-1/2)\sin\frac{2\pi m}{k}.$$

• nicely done (+1) – robjohn Sep 26 '13 at 22:15

In the intro above the question is asked how one might go about computing $$\sum_{n=1}^\infty \frac{1}{3n(3n-1)}.$$

I can contribute an asymptotic expansion and an infinite series to this discussion, shown and proved below. In addition to posting this answer I am also asking two questions I. Can you verify this expansion using a different proof technique? II. Might there even be a reference to the below formula somewhere?

This is the proof, which is procedural, with the result at the end. Suppose we seek to evaluate $$T(q) = \sum_{n\ge 1} \frac{1}{qn(qn-1)} = \sum_{n=1}^p \frac{1}{qn(qn-1)} + \sum_{n\ge 1} \frac{1}{(qn+pq)(qn+pq-1)} \\= \sum_{n=1}^p \frac{1}{qn(qn-1)} + \frac{1}{q^2}\sum_{n\ge 1} \frac{1}{(n+p)(n+(pq-1)/q)}$$ with $p, q\ge 2$ two integers.

The sum term, call it $S(x)$, is harmonic and may be evaluated by inverting its Mellin transform. Recall the Mellin transform identity for harmonic sums with base function $g(x)$, which is $$\mathfrak{M}\left(\sum_{k\ge 1}\lambda_k g(\mu_k x); s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}\right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{(x+p)(x+(pq-1)/q)}.$$ The Mellin transform of $g(x)$ is $$\int_0^\infty \frac{1}{(x+p)(x+(pq-1)/q)} x^{s-1} dx.$$

We chose $p\ge 2$ so that the base function does not have a pole at $x=-1$, which would cause the expansion about infinity that we eventually obtain not to converge. (As we evaluate the harmonic sum at $x=1$ this would be situated right on the boundary between the two disks about zero and about infinity, causing an evaluation at $x=1$ to fail.)

We evaluate the Mellin transform next. Use a circular contour to get $$g^*(s) (1 - e^{2\pi i (s-1)}) \\= 2\pi i \left( \operatorname{Res}\left(\frac{x^{s-1}}{(x+p)(x+\frac{pq-1}{q})}; s=-p\right) + \operatorname{Res}\left(\frac{x^{s-1}}{(x+p)(x+\frac{pq-1}{q})}; s=-\frac{pq-1}{q}\right) \right) \\= 2\pi i \left(-q(-p)^{s-1} + q(-(pq-1)/q)^{s-1}\right) = 2\pi i e^{\pi i (s-1)} q \left(((pq-1)/q)^{s-1}-p^{s-1}\right).$$ This implies that $$g^*(s) = 2\pi i q \frac{-e^{\pi i s}}{1 - e^{2\pi i s}} \left(\left(\frac{pq-1}{q}\right)^{s-1} - p^{s-1}\right)\\= -2\pi i q \frac{1}{e^{-\pi i s} - e^{\pi i s}} \left(\left(\frac{pq-1}{q}\right)^{s-1} - p^{s-1}\right) \\= q\pi\frac{2i}{e^{\pi i s} - e^{-\pi i s}} \left(\left(\frac{pq-1}{q}\right)^{s-1} - p^{s-1}\right) = q\frac{\pi}{\sin(\pi s)} \left(\left(\frac{pq-1}{q}\right)^{s-1} - p^{s-1}\right).$$

It follows that the Mellin transform of the sum term for $S(x)$ is given by $$Q(s) = q \frac{\pi}{\sin(\pi s)} \left(\left(\frac{pq-1}{q}\right)^{s-1} - p^{s-1}\right) \zeta(s) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s).$$

Finally we invert the Mellin transform with the inversion integral $$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ to obtain an expansion about infinity of $S(x)$ starting with the pole at $s=2$ and getting $$S(x) \sim - q \sum_{m\ge 2} (-1)^m \left(\left(\frac{pq-1}{q}\right)^{m-1} - p^{m-1}\right) \frac{\zeta(m)}{x^m}.$$

Setting $x=1$ and substituting into the original equation we finally have $$T(q) = \sum_{n=1}^p \frac{1}{qn(qn-1)} - \frac{1}{q} \sum_{m\ge 2} (-1)^m \left(\left(\frac{pq-1}{q}\right)^{m-1} - p^{m-1}\right) \zeta(m).$$

Addendum Wed Apr 2 20:14:50 CEST 2014. The following MSE link shows how to work with divergent Mellin transforms where there is a pole on the positive real axis (even ones).

• Dear Teacher, I'm so sorry for this comment..I asked a question in MSE and I waited for days..(with the 100+ bountry)..But, unfortunately, I did not get a satisfactory answer...I would like to ask..If You have a few minutes, can you look my problem, Please?.. (I'm sorry for wrong words, English is my second language).. Thank you very much.. math.stackexchange.com/q/2736937/460967 – lone student Jul 7 '18 at 6:52

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}{\left\vert #1\right\rangle} \newcommand{\ol}{\overline{#1}} \newcommand{\pars}{\left(\, #1 \,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}} \newcommand{\verts}{\left\vert\, #1 \,\right\vert} \newcommand{\wt}{\widetilde{#1}}$ \begin{align} &\color{#c00000}{\sum_{n = 2}^{\infty}{\zeta\pars{n} \over k^{n}}} =\sum_{n = 2}^{\infty}{1 \over k^{n}} \sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1}^{n}} =\sum_{\ell = 0}^{\infty}\sum_{n = 2}^{\infty}{1 \over \bracks{k\pars{\ell + 1}}^{n}} =\sum_{\ell = 0}^{\infty} {\bracks{k\pars{\ell + 1}}^{-2} \over 1 - \bracks{k\pars{\ell + 1}}^{-1}} \\[3mm]&={1 \over k^{2}}\sum_{\ell = 0}^{\infty} {1 \over \pars{\ell + 1}\pars{\ell + 1 - 1/k}} ={1 \over k^{2}}\,{\Psi\pars{1} - \Psi\pars{1 - 1/k} \over 1 - \pars{1 - 1/k}} \\[3mm]&=\color{#c00000}{% {1 \over k}\bracks{\Psi\pars{1} - \Psi\pars{1 - {1 \over k}}}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function.

$$\color{#00f}{\large\sum_{n = 2}^{\infty}{\zeta\pars{n} \over k^{n}} =-\,{1 \over k}\,\bracks{\gamma + \Psi\pars{1 - {1 \over k}}}}\,,\qquad\verts{k} > 1$$ since $\ds{\Psi\pars{1} = -\gamma.\quad}$ $\gamma$ is the Euler-Mascheroni Constant.

$\large\tt\mbox{Just one of your questions !!!}$: \begin{align} \sum_{n = 1}^{\infty}{1 \over 3n\pars{3n - 1}}& ={1 \over 9}\sum_{n = 1}^{\infty}{1 \over n\pars{n - 1/3}} ={1 \over 9}\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}\pars{n + 2/3}} ={1 \over 9}\,{\Psi\pars{1} - \Psi\pars{2/3} \over 1 - 2/3} \\[3mm]&={1 \over 3}\,\bracks{\Psi\pars{1} - \Psi\pars{{2 \over 3}}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\pars{1} \end{align} Also,

$$\Psi\pars{{2 \over 3}} = -\gamma + {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}$$ By replacing in $\pars{1}$, we'll find:

$$\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over 3n\pars{3n - 1}} ={1 \over 18}\bracks{9\ln\pars{3} - \root{3}\pi}} \approx 0.2470$$