Some days ago I derived the identity $$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$ where $H_n$ is the $n$th Harmonic number. Other related identities include $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^4}=\frac{-1}{6}\pi^2\zeta(3)+3\zeta(5)$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^5}=\frac{1}{540}(\pi^6-270\zeta(3)^2)$$ (I proved all except the last one) Now I wondered if there is a closed form for the generalized form $$\sum_{n=1}^{\infty}\frac{H_n}{n^x}$$ For some real number $x$. Here is my try: The well known identity $\psi(n+1)=H_n-\gamma$ gives $$\sum_{n=1}^{\infty}\frac{H_n}{n^x}=\sum_{n=1}^{\infty}\frac{\psi(n+1)}{n^x}-\gamma\zeta(x)$$ But I don't know what to do further. Any help would be appreciated.

  • 1
    $\begingroup$ Not so easy: $\sum _{n=1}^{\infty } \frac{H_n}{n^x}=\int_0^{\infty } -\frac{e^t t^{-1+x} \ln\left(1-e^{-t}\right)}{\left(-1+e^t\right) \Gamma (x)} \, dt=\int_0^1 \frac{\text{Li}_x(t)-\zeta (x)}{-1+t} \, dt$ =? $\endgroup$ Dec 13, 2020 at 13:22
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – metamorphy
    Dec 13, 2020 at 13:26
  • $\begingroup$ might be a duplicate. see here math.stackexchange.com/q/469023\ $\endgroup$ Feb 17, 2021 at 4:08

2 Answers 2


Euler proved the following result:

Theorem For integer $q\geq 2$, we have $$\sum_{n=1}^\infty \frac{H_n}{n^q}=\left( 1+\frac{q}{2}\right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1) \zeta(q-k)$$

A proof using the residue theorem can be found in the paper: "Euler sums and contour integral representations" by Philippe Flajolet and Bruno Salvy.


Long Comment:

The sum can be written in terms of unsigned Stirling numbers of the first kind, $s_{n}^{(k)}=\left[_{k}^{n} \right]$

$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n+1}^{(2)} x^n}{n!\; n^{\alpha}}$$

Since $H_n=\left[_{\,\,\,2}^{n+1} \right]\frac{1}{n!}$

The well known recurrence relation for unsigned Stirling numbers of the first kind is

$$\left[_{\,\,\,k}^{n+1} \right]=n\left[_{k}^{n} \right] +\left[_{k-1}^{\,\,n} \right]$$

which immediately leads to

$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n}^{(2)} x^n}{n!\; n^{{\alpha}-1}}+\sum _{n=1}^{\infty } \frac{ x^n}{ n^{{\alpha}+1}}\tag{1}$$

which can (according to Mathematica) be rewritten simply as (with $x=1$)

$$\sum_{n=1}^{\infty}\frac{H_n}{n^{\alpha}}=S_{{\alpha}-1,2}(1)+\text{Li}_{{\alpha}+1}(1)=S_{{\alpha}-1,2}(1)+\zeta({\alpha}+1)$$ where $S_{n,p}(z)=\frac{(-1)^{n+p-1}}{(n-1)! p!}\int _0^1\frac{ t \log ^{n-1} \left(\log ^p (1-t z)\right)}{t} \,dt$ is the Nielsen generalized polylogarithm function and $\text{Li}_n$ is the polylogarithm function.

[Update: The full infinite series form for the Nielsen generalized polylogarithm according to Mathematica is

$$S_{n,p}(z)=\sum _{k=1}^{\infty } \frac{ s_k^{(p)}}{k! \,k^n}z^k$$

However I cannot find this result online.]

The Nielsen generalized polylogarithm function is utilised in the study of quantum electrodynamics according to information available online, so its properties have been quite well studied apparently. I didn't have time to look further, but hopefully this gives one potential starting point in your quest for an answer.


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