2
$\begingroup$

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 $p=2,3,\dots?$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.