Thursday, January 2, 2020

sequences and series - Closed form expressions for harmonic sums



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?


Answer



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.



Derivation




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]

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