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 $-1\leq z \leq 1$?
