# On generalizing the harmonic sum $\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n = S_{k-1,2}(1)+\zeta(k+1)$ when $z=1$?

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 I have enumerated $F_k(+1)$ for $k=1,\cdots,11$ and also got some results on $F_k(z)$ in math.stackexchange.com/questions/2169507/… . 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? 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. May 31 '19 at 14:16

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.

• The case when $z=\frac12$ for $F_2(z)$ and $F_3(z)$ is discussed in this and this post. Jun 1 '19 at 6:37
• 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...) Jun 1 '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. Jun 3 '19 at 15:49