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?