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

Question

‎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 $\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$?

Answer 1

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 $$ H_{-j/m}-H_{-(m-j)/m}=\pi\cot\left(\frac{\pi j}{m}\right)\tag{6} $$

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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, $$ \begin{align} \sum_{n=1}^\infty\frac1{3n(3n-1)} &=-\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/3}\right)\\ &=-\frac13H_{-1/3}\\[6pt] &=\frac12\log(3)-\frac\pi{6\sqrt3}\tag{10} \end{align} $$

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Values for Future Reference

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

Answer 2

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 $$\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}=\frac{1}{6} \left(3\log(3) -\frac{\pi }{\sqrt{3}}\right)$$

Answer 3

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 $$\sum_n\frac{1}{kn(kn-1)}=\frac{1}{k}\sum_{\alpha^k=1,\alpha\neq1}(1-\alpha)\log(1-\alpha)=$$ ($\alpha=\exp(2\pi i m/k)$) $$=\frac{1}{k}\sum_{m=1}^{k-1}((1-\cos\frac{2\pi m}{k})-i\sin\frac{2\pi m}{k})(\log(2\sin\frac{\pi m}{k})+\pi i(m/k-1/2))=$$ $$=\frac{1}{k}\sum_{m=1}^{k-1}(1-\cos\frac{2\pi m}{k})\log(2\sin\frac{\pi m}{k})+\pi(m/k-1/2)\sin\frac{2\pi m}{k}.$$

Answer 4

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

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

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

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

Answer 5

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).