Motivated by analytic continuation of solutions of a Picard-Fuchs equation, we encountered sums of the following form


$S(z;p)=\sum\_{k=1}^{\infty}(-1)^{k+1} (H\_k)^p z^k$


where $H\_k = \sum\_{n=1}^{k} 1/n$ are the harmonic numbers and $p \in \mathbb{N}.$


For $p=1$, $S(z;1) = \frac{\log(1+z)}{1+z}$ and at $z=1$, $S(z=1;1)=1/2 \log 2$. In general, we expect that the sums for $p>1$ will have a radius of convergence 1, with a pole at $z=-1$, but that the analytic continuation to $z=1$ will have a finite value. This value is expected to be a sum of multiple zeta functions. Are these sums known for $p=2,3,\dots?$



Alternating series $\sum_{k=1}^{\infty}(-1)^{k+1} (H_k)^p z^k$ and multiple zeta values
If we multiply $S(z;p)$ by $1+z$, we get
\begin{align\*}
(1+z)S(z;p)&=\sum\_{k\geq 1}(-1)^{k+1}\left(H\_k^p-H\_{k-1}^p\right)z^k\\
&=\sum\_{k\geq 1}(-1)^{k+1}\sum\_{n=0}^{p-1}{p\choose n}H\_{k-1}^n\frac{1}{k^{p-n}}z^k\\
&=\sum\_{n=0}^{p-1}{p\choose n}\sum\_{k\geq 1}\frac{(-1)^{k+1}H\_{k-1}^n}{k^{p-n}}z^k.
\end{align\*}
These computations are valid for $|z|<1$, but this last series actually converges at $z=1$. Using the "stuffle product" to expand out the terms $H\_{k-1}^n$, we can write $S(1;p)$ as a linear combination of alternating Euler sums (which are like multiple zeta values except one allows an alternating sign). There are a bunch of known relations among the alternating Euler sums, but it's not clear to me whether it will be possible to use these relations to write $S(1;p)$ in terms of (non-alternating) multiple zeta values.


