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The following atypical harmonic series was recently proposed by C.I. Valean, $$ \sum_{n=1}^{\infty}\frac{H_n H_{4n}}{n^2}$$ $$=\frac{13}{12}\log^4(2)-\frac{35}{4}\log^2(2)\zeta(2)+\frac{91}{4}\log(2)\zeta(3)-\frac{665}{16}\zeta(4)+6 \log(2)\pi G -8 G^2$$ $$+12 \pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}+26 \operatorname{Li}_4\left(\frac{1}{2}\right),$$

where $\zeta$ is the Riemann zeta function, $G$ represents the Catalan's constant, and $\operatorname{Li}_n$ denotes the Polylogarithm.

His solution starts with using the well-known property with harmonic numbers and skew-harmonic numbers, that is $\overline{H}_{2n}=H_{2n}-H_n$, or $H_n=H_{2n}-\overline{H}_{2n}$, which leads to \begin{equation*} \sum_{n=1}^{\infty}\frac{H_n H_{4n}}{n^2}=\sum_{n=1}^{\infty}\frac{(H_{2n}-\overline{H}_{2n}) H_{4n}}{n^2}=4\sum_{n=1}^{\infty} \frac{H_{2n} H_{4n}}{(2n)^2}-4\sum_{n=1}^{\infty} \frac{\overline{H}_{2n} H_{4n}}{(2n)^2} \end{equation*} \begin{equation*} =2\sum_{n=1}^{\infty} \frac{H_n H_{2n}}{n^2}-2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n H_{2n}}{n^2}-2\sum_{n=1}^{\infty} \frac{\overline{H}_n H_{2n}}{n^2}+2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{\overline{H}_n H_{2n}}{n^2}, \end{equation*} where the series $$\sum _{n=1}^{\infty} \frac{\overline{H}_nH_{2n}}{n^2}=\frac{507}{64}\zeta(4)-\frac{7}{4}\log(2)\zeta(3)+\frac{5}{4}\log^2(2)\zeta(2)-\frac{7}{48}\log^4(2)-2\pi \Im\biggr \{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr \}-\frac{7}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)$$

and

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{\overline{H}_n H_{2n}}{n^2}=\frac{5}{48}\log^4(2)-\frac{5}{8}\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{77}{32}\zeta(4)+\log(2)\pi G-2 G^2+\frac{5}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)$$

are found in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), Chapter $4$, Section $4.57$, p.$454$, then $$ \sum_{n=1}^{\infty}\frac{H_n H_{2n}}{n^2}=\frac{13}{8}\zeta(4)+\frac{7}{2}\log(2)\zeta(3)-\log^2(2)\zeta(2)+\frac{1}{6}\log^4(2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right)$$

is given in the same book, Chapter $4$, Section $4.23$, p.$428$, and finally $$ \sum _{n=1}^{\infty} (-1)^{n-1} \frac{H_n H_{2n}}{n^2}=2G^2-2\log(2)\pi G-\frac{1}{8}\log^4(2)-\frac{21}{8}\log(2)\zeta(3)+\frac{3}{2}\log^2(2)\zeta(2)+\frac{773}{64}\zeta(4)-4\pi \Im\biggr \{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr \}-3\operatorname{Li}_4\left(\frac{1}{2}\right),$$ presented in the same reference, Chapter $4$, Section $4.55$, p.$453$, thus giving the desired result.

Questions:

$1)$. Are such results known in the literature, or similar ones like, say, $\displaystyle \sum_{n=1}^{\infty}\frac{H_n H_{4n}}{(2n-1)^2}$?

$2).$ I would love to see very different ideas, and strategies leading to the desired result, and at the same time remaining in the realm of simple real methods. So, what other ways would we like to explore? (there is absolutely no hurry)

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    $\begingroup$ \begin{align}\sum_{n=1}^{\infty}\frac{H_n H_{4n}}{n^2}=-\int_0^1\int_0^1 \frac{\ln(1-x)\ln\left(\frac{1-xy^4}{1-x}\right)}{x(1-y)}dxdy\end{align} $\endgroup$
    – FDP
    Aug 29, 2023 at 13:41
  • $\begingroup$ @FDP Nice double integral representation. $\endgroup$ Aug 29, 2023 at 19:21
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    $\begingroup$ I'm trying to compute it. $\endgroup$
    – FDP
    Aug 29, 2023 at 21:01

1 Answer 1

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A very long comment, definitely not a "simple real method" that OP is looking for.

The structure of such sums in terms of multiple zeta values (MZVs) are considered well-understood. Several years ago I implemented an algorithm to calculate logarithmic integrals via MZVs, the algorithm for Euler sums is almost the same. Now you can compute them using the same Mathematica package.

An overview of the theory and algorithm behind can be learned here.


For OP's sum, one simply input enter image description here the program only accepts sum from $0$ to $\infty$, so we need to do a shift by $1$. This corroborates OP's result.

What OP called "skew-harmonic number" $\overline{H}_n = \sum_{i=1}^n \frac{(-1)^{i-1}}{i}$ is represented in the program as MultiHarmonic[{1},{1},{-1},n], the two sums OP mentioned can also be computed using same algorithm:

enter image description here

OP asks about the sum $\sum_{n=1}^{\infty}\frac{H_n H_{4n}}{(2n-1)^2}$ at the end, it equals enter image description here note that the weight is now inhomogeneous.


All variations of Euler sums in the literature can be routinely converted to MZVs. More examples that OP might enjoy solving using "simple real method" are: $$\sum_{n=1}^\infty \frac{(H_n)^7}{n^2} \qquad \sum_{n=1}^\infty \frac{H_n H_n^{(2)} H_n^{(3)}}{n^3}\qquad \sum_{n=1}^\infty \frac{(-1)^n \overline{H}_n \overline{H}_n^{(2)} H_n^{(2)}}{n}$$

$$\sum_{n=0}^\infty \frac{H_n H_{2n}}{(4n+1)^2} \qquad \sum_{n=0}^\infty \frac{H_n H_{4n}}{(4n+1)^2}\qquad \sum_{n=0}^\infty \frac{H_{2n} H_{4n}}{(4n+1)^2}$$ the middle sum, for example, is $$-\frac{\pi ^2 G}{32}+\frac{7}{8} G \log ^2(2)+\frac{\pi}{8} G \log (2)-\beta(4)+\frac{\pi}{4} \Im \text{Li}_3(\frac{1+i}{2})+\Im\text{Li}_4(\frac{1+i}{2})+2 \log (2) \Im \text{Li}_3(\frac{1+i}{2})+\frac{3 \text{Li}_4\left(\frac{1}{2}\right)}{2}+\frac{7 \pi \zeta (3)}{64}+\frac{21}{32} \zeta (3) \log (2)-\frac{53 \pi ^4}{3840}+\frac{\log ^4(2)}{16}-\frac{11}{192} \pi \log ^3(2)-\frac{9}{128} \pi ^2 \log ^2(2)-\frac{9}{256} \pi ^3 \log (2)$$

as well as $$\sum_{n\geq 0} \frac{(-1)^n A_n H_n^{(2)} H_n}{2n+1} \qquad \sum_{n\geq 0} \frac{(-1)^n A_n H_n^{(2)}}{n+1} \qquad A_n = \sum_{i=0}^{n-1} \frac{(-1)^{i-1}}{2i-1}$$

Similar to logarithmic integrals, one can literally continue forever: there are millions if not billions of variations, they can however all be tackled using the same method.


Nonetheless, there are still much room of improvement for the program: currently it can't

  1. reduce sums such as $\sum_{n\geq 1} \frac{H_n^{(p)}}{n^q}$ with $p+q$ odd using $\zeta(s)$, this is well-known to be possible (parity-depth reduction of MZV)
  2. handle generalizations of Euler sums that contain binomial coefficient such as this one.

Algorithms for both points are understood, and will be implemented in future.

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    $\begingroup$ (+1) Thanks for sharing. $\endgroup$ Aug 31, 2023 at 7:59
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    $\begingroup$ @user97357329 Yes. i.stack.imgur.com/Nt0xG.png, we can even evaluate each of them. $\endgroup$
    – pisco
    Aug 31, 2023 at 17:13
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    $\begingroup$ @user97357329 thank you very much for the appreciation. $\endgroup$
    – pisco
    Aug 31, 2023 at 19:20
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    $\begingroup$ The previous series could have been arranged way more and simplified, but I left it that way just as an example. Much, much more complicated series alike are possible. $\endgroup$ Aug 31, 2023 at 19:34
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    $\begingroup$ @user97357329 Indeed a very curious result. I look forward to you sharing more of your beautiful results on this site. $\endgroup$
    – pisco
    Aug 31, 2023 at 20:18

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