# Closed form for $f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n$?

So I have been working with some power series involving the Harmonic numbers. I have been able to evaluate the first couple of sums as $$f(z,0)=\sum_{n=1}^\infty H_n z^n=\sum_{n=1}^\infty z^n\int_0^1 \frac{1-x^n}{1-x}dx=-\frac{\ln(1-z)}{1-z}$$ $$f(z,1)=\sum_{n=1}^\infty \frac{H_n}{n}z^n=\int_0^z\frac{f(x,0)}{x}dx=\operatorname{Li}_2(z)+\frac{\ln^2(1-z)}{2}$$ $$f(z,2)=\sum_{n=1}^\infty \frac{H_n}{n^2}z^n=\int_0^z \frac{f(x,1)}{x}dx=\operatorname{Li}_3(z)-\operatorname{Li}_3(1-z)+\operatorname{Li}_2(1-z)\ln(1-z)+\frac{\ln(z)\ln^2(1-z)}{2}+\zeta(3)$$ and in general$$f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n=\int_0^z \frac{f(x,a-1)}{x}dx$$ I was wondering if it is possible to derive a general formula for $$f(z,a)$$ as computing each integral of $$\frac{f(x,a-1)}{x}$$ becomes very tedious and complicated to do and does not seem to provide a closed form answer for any $$a$$. I know that there exists a closed form for $$f(1,a)$$ but how could we generalize this to $$f(z,a)$$?

• Is there any condition on $\displaystyle a$ ?. – Felix Marin May 31 '19 at 23:06
• @FelixMarin I'm looking for the case where $a$ is a positive integer but it would also be nice to know a formula for non-integer $a$. – aleden May 31 '19 at 23:10

$$F_a(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^a}z^n= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$
with polylogarithm $$\rm{Li}_{a+1}(z)$$ and Nielsen generalized polylogarithm $$S_{n,p}(z)$$. The situation then is subsumed by the Nielsen polylogs.
There are formulas for general $$z$$ when $$a=1,2$$ (as you mentioned) as well as for $$a=3$$, though I am not sure if there is for higher.
• See this post for $a=3$. – Tito Piezas III Jun 8 '19 at 3:36
• For the special cases $z=\frac12$ and $z=-1$, closed-forms are discussed here and here. – Tito Piezas III Jun 8 '19 at 4:16