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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 $f(z,a)$?

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  • $\begingroup$ Is there any condition on $\displaystyle a$ ?. $\endgroup$ – Felix Marin May 31 '19 at 23:06
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    $\begingroup$ @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 $a$. $\endgroup$ – aleden May 31 '19 at 23:10
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By coincidence, I asked virtually the same question. I managed to answer it myself which is,

$$F_a(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^a}z^n= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$

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.

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  • $\begingroup$ See this post for $a=3$. $\endgroup$ – Tito Piezas III Jun 8 '19 at 3:36
  • $\begingroup$ For the special cases $z=\frac12$ and $z=-1$, closed-forms are discussed here and here. $\endgroup$ – Tito Piezas III Jun 8 '19 at 4:16

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