It is well known that there are deep connections between harmonic sums (discrete infinite sums that involve generalized harmonic numbers) and poly-logarithms. Bearing this in mind we have calculated the following sum: \begin{equation} S_a(t) := \sum\limits_{m=1 \vee a} H_m \cdot \frac{t^{m+1-a}}{m+1-a} \end{equation} where $t\in (-1,1)$ and $a \in {\mathbb Z}$. The result reads: \begin{eqnarray} S_a(t) = \left\{ \begin{array}{lll} \frac{1}{2} [\log(1-t)]^2 + \sum\limits_{j=1}^{a-1} \frac{1}{j \cdot t^j} \left( \sum\limits_{m=1}^j \frac{t^m}{m} + (1-t^j) \log(1-t) \right) + Li_2(t) 1_{a\ge 1} & \mbox{if $a \ge 0$} \\ \frac{1}{2} [\log(1-t)]^2 -\sum\limits_{j=1}^{|a|} \frac{1}{j } \left( \sum\limits_{m=1}^j \frac{t^m}{m} + (1-t^j) \log(1-t) \right) & \mbox{if $a < 0$} \\ \end{array} \right. \end{eqnarray} Unfortunately it took me a lot of time to derive and thoroughly check the result even though all the calculations are at elementary level. It is always helpful to use Mathematica. Indeed for particular values of $a$ Mathematica "after long thinking" comes up with solutions however from that it is hard to find the generic result as given above. Besides in more complicated cases Mathematica just fails.

In view of the above my question is the following. Can we prove that every infinite sum whose coefficients represent a rational function in $m$ and in addition involve products of positive powers of generalized harmonic numbers, that such a sum is always always given in closed form by means of elementary functions and poly-logarithms? If this is not the case can we give a counterexample?

  • $\begingroup$ Hm, what a general question. Perhaps you could write out the general form you are looking for to avoid possible misinterpretations? $\endgroup$ – Simply Beautiful Art Apr 19 '17 at 12:01
  • $\begingroup$ Since every rational function in the variable $m$ can be decomposed into partial fractions I guess the general form would be \begin{equation} \sum\limits_{m=1\vee a}^\infty H_m^{(n_1)} \cdot \frac{t^{m+1-a}}{\left(m+1-a\right)^{n_2}}\end{equation} where $n_1$ and $n_2$ are some strictly positive integers. $\endgroup$ – Przemo Apr 19 '17 at 14:00

Even though this is not conceived as an answer to your specific question which conserns the class of functions needed to represent your sum it also contributes to it as it exhibits a broader class than you mentioned.

Also I think it is an interesting result in itself when it comes to closed formes.

Compact closed expression

I have found that your sum

$$S_a(t) := \sum\limits_{m=1} ^{ \infty} H_m \cdot \frac{t^{m+1-a}}{m+1-a}\tag{1}$$

can be expressed in more compact closed forms. The first one I derived is the following; for the second one see the paragraph "Derivation".

$$f(a,t) = -\frac{1}{12} \left(12 a \, _4F_3(1,1,1,a+1;2,2,2;1-t)-12 a t \, _4F_3(1,1,1,a+1;2,2,2;1-t)-12 a \log (1-t) \, _3F_2(1,1,a+1;2,2;1-t)+12 a t \log (1-t) \, _3F_2(1,1,a+1;2,2;1-t)+6 \psi ^{(0)}(1-a)^2+12 \gamma \psi ^{(0)}(1-a)-6 \psi ^{(1)}(1-a)+6 \gamma ^2+\pi ^2\right)+\frac{1}{2} \log ^2(1-t)$$

It consists of hypergeometric, polygamma and log functions, and some well known constants abundant in this field.

The graph shows $f(a,t=1/2)$ as a function of $a$

enter image description here

Validity checks

I have checked the validity of both $f(a,t)$ and $f_{1}(a,t)$ by plotting them together with a partial sum of $S_a(t)$ as a function of $a$ for $t=1/2$. All three curves agree reasonably.

Unfortunately, my attempts to perform an independent validity check by comparing power series in $t$ failed. This might be due to difficulties in Mathematica and needs further study.


We observe that the derivative of $S_a(t)$ with respect to $t$ is a simple function

$$\frac{\partial S_a(t)}{\partial t}=\sum _{m=1}^{\infty } H_m t^{m-a}=-\frac{t^{-a} \log (1-t)}{1-t}\tag{2}$$

Hence $S_a(t)$ is given by the integral

$$-\int_0^t \frac{u^{-a} \log (1-u)}{1-u} \, du\tag{3}$$

Mathematica gives for this integral the following expression

$$f_{1}(a,t) =\pi (-1)^{a-1} H_{a-1} \csc (\pi a)-\frac{1}{a^2}\left(\frac{1}{t-1}\right)^a \, _3F_2\left(a,a,a;a+1,a+1;\frac{1}{1-t}\right)-\frac{1}{a}\left(\frac{1}{t-1}\right)^a \log (1-t) \, _2F_1\left(a,a;a+1;\frac{1}{1-t}\right)$$

This is equivalent to $f(a,t)$ which I have derived first in the following more complicated manner.

Substituting $u\to 1-s$ in $(3)$ leads to

$$-\int_{1-t}^1 \frac{(1-s)^{-a} \log (s)}{s} \, ds\tag{4}$$

Expanding $(1-s)^{-a}$ into a binomial series, interchanging the summation and integration, doing the integral, and then the sum gives $f(a,t)$.

Mathematica expressions

In order to avoid possible typing errors, here are the Mathematica expressions for the functions obtained

f[a_, t_] := (1/
    2 Log[1 - t]^2 + -(1/
     12) (6 EulerGamma^2 + \[Pi]^2 + 
      12 a HypergeometricPFQ[{1, 1, 1, 1 + a}, {2, 2, 2}, 1 - t] - 
      12 a t HypergeometricPFQ[{1, 1, 1, 1 + a}, {2, 2, 2}, 1 - t] - 
      12 a HypergeometricPFQ[{1, 1, 1 + a}, {2, 2}, 1 - t] Log[
        1 - t] + 
      12 a t HypergeometricPFQ[{1, 1, 1 + a}, {2, 2}, 1 - t] Log[
        1 - t] + 12 EulerGamma PolyGamma[0, 1 - a] + 
      6 PolyGamma[0, 1 - a]^2 - 6 PolyGamma[1, 1 - a]))

 f1[a_, t_] :=(-1)^(a - 1)  \[Pi] Csc[a \[Pi]] HarmonicNumber[a - 1] - (1/
   a^2) (1/(-1 + t))^
  a HypergeometricPFQ[{a, a, a}, {1 + a, 1 + a}, 1/(1 - t)] - (1/
   a) (1/(-1 + t))^
  a Hypergeometric2F1[a, a, 1 + a, 1/(1 - t)] Log[1 - t]

Let us firstly set $a=0$. Even in this is not going to be a full answer to this question and I don't even pretend that I will provide the answer whether poly-logarithms and elementary functions suffice in order to evaluate the sums above or whether something more is required. Yet since harmonic sums attracted a lot of interest in this website and also since I need to re-sum harmonic series for my own purposes I just want to write out some result that might be a starting point for investigating my question.If we just start from the generating function of generalized harmonic numbers and then subsequently divide the right hand side by $t$ and integrate and repeat that operation several times then we easily get the following result: \begin{equation} S_{a,n_1}^{(n)}:=\sum\limits_{m=1}^\infty H_m^{(n)} \cdot \frac{t^{m+1}}{(m+1)^{n_1}} = \int\limits_0^t \frac{\left( \log(\frac{t}{\eta}\right)^{n_1-1}}{(n_1-1)!} \cdot \frac{Li_n (\eta)}{1-\eta} d\eta \end{equation} where $n$ and $n_1$ are strictly positive integers and $t \in (-1,1)$.

Now, in particular case it actually appears that the quantities above do indeed reduce to elementary functions and poly-logarithms only. For example we have: \begin{eqnarray} S_{a,2}^{(1)} &=& \frac{1}{2} \log(t) \cdot [\log(1-t)]^2 + Li_2(1-t) \cdot \log(1-t) + Li_3(1) - Li_3(1-t) \\ S_{a,1}^{(2)} &=& -\frac{\pi^2}{3} \log(1-t) + \log(t) [\log(1-t)]^2 + \log(1-t) Li_2(t) + 2 Li_3(1-t) - 2 Li_3(1) \\ S_{a,1}^{(3)} &=& -\log(1-t) Li_3(t) - 1/2 [Li_2(t) ]^2 \end{eqnarray} which we derived from the formula above using integrating by parts.

Unfortunately we were not able to compute $S^{(1)}_{a,3}$ with this method. Mathematica also fails in this case. We can only say --again from integrating by parts once-- that : \begin{equation} \sum\limits_{m=1}^\infty H_m^{(1)} \cdot \frac{t^{m+1}}{(m+1)^n} = PolyLog[n-1,2,t] \end{equation} for $n\ge 2$. Here $PolyLog[\cdot,\cdot,t]$ is the Nielsen generalized poly-logarithm. It is unclear whether that Nielsen-quantity reduces to polylogarithms only or whether Nielsen introduced that quantity simply because he wasn't able to resolve the question we are posing. Therefore we haven't answered our original question yet we could sligthly change the question we are asking namely what harmonic sums are the Nielsen-generalized poly-logarithms related to. This question is easily answered. Indeed, by using the Taylor expansion of $[\log(1-\eta)]^q$ (see answer to the question Compute an integral containing a product of powers of logarithms.) and by using the relationships between Stirling numbers of the first kind and generalized harmonic numbers (see https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) it is pretty straightforward to show that: \begin{eqnarray} \sum\limits_{m=1}^\infty \left([H_m^{(1)}]^2 - H_m^{(2)}\right)\cdot \frac{t^{m+1}}{(m+1)^n} &=& 2 PolyLog[n-1,3,t] \\ \sum\limits_{m=1}^\infty \left([H_m^{(1)}]^3 - 3 H_m^{(2)} H_m^{(1)} + 2 H_m^{(3)}\right) \cdot \frac{t^{m+1}}{(m+1)^n} &=& 6 PolyLog[n-1,4,t] \\ \vdots \\ \sum\limits_{m=1}^\infty \left[\begin{array}{r} m+1 \\ q\end{array}\right] \cdot \frac{(q-1)!}{(m)!} \cdot \frac{t^{m+1}}{(m+1)^n} &=& (q-1)! PolyLog[n-1,q,t] \end{eqnarray} where $\left[\begin{array}{r} m+1 \\ q\end{array}\right]$ are the unsigned Stirling numbers of the first kind. In here $n \ge 2$ and $q=1,2,3,\cdots$. Finally, by using the result in Compute an integral containing a product of powers of logarithms. by using the symmetry of the poly-logarithm in its first two variables we get the following identities: \begin{eqnarray} &&\sum\limits_{m=1}^\infty \frac{H_m}{(m+1)^n} = \frac{1}{2} \cdot \left(n \zeta(1+n) - \sum\limits_{j=1}^{n-2} \zeta(j+1) \zeta(n-j)\right) \\ &&\sum\limits_{m=1}^\infty \frac{H_m^2 - H_m^{(2)}}{(m+1)^n} =\\ && \frac{1}{3!} \left( 2 n^{(2)} \zeta(n+2) - 3 \sum\limits_{j=1}^{n-2} \left((j+1) \zeta(j+2) \zeta(n-j) + (n-j) \zeta(j+1) \zeta(n-j+1) \right) + 2 \sum\limits_{1 \le j < j_1 \le n-2} \zeta(j+1)\zeta(j_1-j+1)\zeta(n-j_1) \right) \end{eqnarray} where $\zeta()$ is the zeta function.

Having said all this I reiterate the original question . Do all the harmonic sums in question indeed reduce to elementary functions and polylogarithms only or instead some more general functions are needed to get closed forms.

  • $\begingroup$ +1 Thanks for this answer. It enabled to see a relation between these harmonic sums and the Nielsen generalized polylogarithm. Kindly see this post $\endgroup$ – Tito Piezas III May 21 at 14:58
  • $\begingroup$ Maybe you can shed some light on this new post? $\endgroup$ – Tito Piezas III May 31 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.