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Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$. In this post it asks for the evaluation,

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\tfrac{5}{4}\zeta(4)$$

while this post and this answer discusses the next one,

$$ \sum_{n=1}^{\infty}\frac{H_n}{n^4} = -\zeta(2)\zeta(3)+3\zeta(5)$$

Given the more general sum,

$$F_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n\tag1$$

It seems the special cases $z=1$ is,

$$F_k(1)= \sum_{n=1}^{\infty}\frac{H_n}{n^k} = S_{k-1,2}(1)+\zeta(k+1)\tag2$$

while $z=-1$ is,

$$F_k(-1)= \sum_{n=1}^{\infty}\frac{H_n}{n^k}(-1)^n = S_{k-1,2}(-1)-\frac{2^k-1}{2^k}\zeta(k+1)\tag3$$

where $S_{n,p}(z)$ is the Nielsen generalized polylogarithm,

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

However, for the range $-1\leq z \leq 1$, the closely related sum in this answer has a simple formula,

$$G_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^k}\,z^{n+1} = S_{k-1,2}(z)\tag4$$

Q: Like $G_k(z)$, does $F_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n$ have a common closed-form in the range $-1\leq z \leq 1$?

Tito Piezas III
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  • @ Tito Piezas III I have enumerated $F_k(+1)$ for $k=1,\cdots,11$ and also got some results on $F_k(z)$ in https://math.stackexchange.com/questions/2169507/calculating-alternating-euler-sums-of-odd-powers/2264045#2264045 . I guess you ponder on what is the relationship between Nielsen poly-logs and poly-logs. In general I do not know that. I lack motivation to further investigate those things. Can i ask you why are you interested in those things? – Przemo May 31 '19 at 11:01
  • @Przemo:I must confess it is pure curiosity. I came across the Nielsen polylogs while trying to solve this log sine integral. – Tito Piezas III May 31 '19 at 14:16

1 Answers1

3

After my last edit, I figured out a way to partially answer my question. The trick is to test,

$$F_k(z) - G_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n - \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^k}z^{n+1}$$

for various values of $k,z$ to see if it yields something familiar. For $k=2$ and value $z = 1/3$, the Inverse Symbolic Calculator was able to recognize it as,

$$F_2\big(\tfrac13\big) - G_2\big(\tfrac13\big) = \rm{Li}_3\big(\tfrac13\big)$$

A little more testing showed that for $-1\leq z\leq 1$, apparently,

$$F_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + S_{k,1}(z)$$

with Nielsen generalized polylogarithm $S_{n,p}(z)$. Equivalently, in terms of polylogarithm $\rm{Li}_n(z)$,

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

For the special cases $z=1$ and $z=-1$, the polylogarithm reduces to formulas $(2)$ and $(3)$ in the post.

P.S. Of course, what remains is to rigorously prove the proposed formula.

Tito Piezas III
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  • The case when $z=\frac12$ for $F_2(z)$ and $F_3(z)$ is discussed in this and this post. – Tito Piezas III Jun 01 '19 at 06:37
  • Hmm in fact you simply noticed that $$\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^k}z^{n+1}=\sum_{m=2}^{\infty}\frac{H_{m-1}}{m^k}z^m=\sum_{m=2}^{\infty}\frac{H_{m}-\frac 1m}{m^k}z^m=\sum_{m=1}^{\infty}\frac{H_{m}}{m^k}z^m-\sum_{m=1}^{\infty}\frac 1{m^{k+1}}z^m$$ (sorry...) – Raymond Manzoni Jun 01 '19 at 18:52
  • @RaymondManzoni: No problem. The important thing is the observation is correct. Furthermore, it makes explicit the connection between those harmonic sums and the Nielsen polylogs. – Tito Piezas III Jun 03 '19 at 15:49