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For the past couple of days I have been looking at Euler Sums, and I happened upon this particular one: $$ \sum_{n=1}^{\infty}\left(-1\right)^{n}\, \frac{H_{n}}{n^{5}} $$ I think most people realize why this one is so significant, as of writing this it currently has no closed form. This post is less about finding one, and more about asking why this is the case, as well as exploring the different strategies I have seen that have been used for the lower order cases.
$$ \sum_{n = 1}^{\infty}\left(-1\right)^{n}\, \frac{H_{n}}{n^{3}} $$ This particular sum has a well known result, and I have seen it tackled in a couple different ways. Here is one method below.

Inverse Trigonometric Integrals

What was interesting to me is that within the answer of this post, they also managed to derive a generating function for $\sum_{n=1}^{\infty}x^{n}H_{n}/n^{3}$, and they managed to do it without needing the generating function of, $\sum_{n = 1}^{\infty}x^{n}H_{n}/n^{2}$. So that begs the question, is it possible to create generating functions for the higher order cases, namely $$ \sum_{n = 1}^{\infty}\frac{H_{n}}{n^{4}}\,x^{n},\qquad \sum_{n = 1}^{\infty}\frac{H_{n}}{n^{5}}x^{n} $$

  • Just looking at the monstrosity that exists for the cubed case, the higher order sums feel like their generating functions would be abominations.
  • But my question in this case persists as the following: Can we find those generating functions in a closed form expressed in terms of Polylogarithms and Logarithms ?.
  • Or would we eventually get stuck with an integral of an expression that cannot be dealt with easily ?.
  • It looks like there was some work done in the post below with regards to the $\mbox{"$\,n^{4}\,$" case}$, but there was some debate about its validity.

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

  • Now here is another post that I found while doing some research on this topic: Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^5 2^n}$

  • Now I really do not have a background in Algebra and a lot of the nomenclature and notation is very unfamiliar to me, but I can at least recognize the idea that they arrived at a possible answer to what the answer of the sum could be ?.

  • Could anyone help me understand how this result was achieved, as well its validity.

  • It is my understanding that the result was numerically found, but that no formal proof exists ?.

  • I will link the paper that is referenced also for people who want to read more.

https://arxiv.org/pdf/1908.04770

If anyone has any other discoveries, as well as possible links to any papers that have tackled this series or any other series with associated Weight 6, please feel free to mention or link them.

Felix Marin
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Jessie Christian
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1 Answers1

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A brief answer, that I will try to update if I have more time in the future. In the paper linked https://arxiv.org/pdf/1908.04770 Equation (19) gives an evaluation for $$ \DeclareMathOperator{\Li}{Li} \sum_{n=1}^\infty \frac{H_{n-1} \cdot (-1)^n}{n^5} \,. $$ (You can show that $S_{4,2}(x) = \sum_{n=1}^\infty \frac{H_{n-1}}{n^5} x^n$, or check the first few terms in Mathematica, using Series[PolyLog[4, 2, x], {x, 0, 10}], say. The function $S_{4,2}$ is one of Nielsen's generalised polylogarithms https://mathworld.wolfram.com/NielsenGeneralizedPolylogarithm.html You could write it via a depth 2 mutliple polylogarithm, but there is provably no way to write it via depth 1 functions for generic $x$.)

Since $ H_n = \frac{1}{n} + H_{n-1} $, you can get an evaluation for $$ \sum_{n=1}^\infty \frac{H_{n-1} \cdot (-1)^n}{n^5} \,, $$ just by adding $ \sum_{n=1}^\infty \frac{(-1)^n}{n^6} = -\frac{31}{32} \zeta(6) $.

In particular: \begin{align*} \sum_{n=1}^\infty\frac{(-1)^n H_n}{n^5} = {} & S_{4,2}(-1) - \frac{31}{32} \zeta(6) = \zeta(\overline{5}, 1) - \frac{31}{32} \zeta(6) \\ = {} & \frac{1}{13} \bigg(\frac{1}{3}\Li_6\Big(-\frac{1}{8}\Big)-162 \Li_6\Big(-\frac{1}{2}\Big)-126 \Li_6\Big(\frac{1}{2}\Big)\bigg) -\frac{4783 }{1248}\zeta (6) +\frac{3}{8} \zeta (3)^2 \\ & {}+\frac{31}{16} \zeta (5) \log(2) -\frac{15}{26} \zeta (4) \log ^2(2) +\frac{3}{104} \zeta (2) \log^4(2) -\frac{1}{208} \log ^6(2) \,, \end{align*}

[This was already stated directly in the answer to the other question you linked to, https://math.stackexchange.com/a/3415727/112256 ]

In any case you may object that in the arXiv version we stated this as a numerically checked identity, and gave some motivic reason that it should be true (up to not fixing the rational in front of $\zeta(6)$). We ought to update the arXiv version, because there were significant additions which went into the published version https://dx.doi.org/10.4310/CNTP.2021.v15.n2.a4 ; in particular we found a weight 6 multiple polylogarithm functional equations which allows one to derive the equation (19) as a specialisation. [It's a significant amount of computer algebra with a >100-term long identity, so it would still be nice to find a more conceptual/direct/elementary proof. But there is a closed form evaluation in any case.]

Steven Charlton
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