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 $p=2,3,\dots?$

  • $\begingroup$ By approximating $H_k$ with $\ln k$, I get an expression involving p-th order partial derivatives of the polylogarithmic and Lerch $\Phi$ functions. $\endgroup$ – Lucian Apr 1 '14 at 13:25

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.


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