The Euler sums are given by $$S_{p,q} = \sum_{n = 1}^{\infty} \frac{H_{n}^{(p)}}{n^q},$$ where $$H_{n}^{(p)} = \sum_{j = 1}^{n} \frac{1}{j^p}.$$ According to Wolfram, Eq. (19), the following special case holds: $$S_{1,3} \equiv \sum_{n = 1}^{\infty} \frac{H_{n}}{n^3} = \frac{5}{4}\zeta(4),$$ where $H_n = H_n^{(1)}$ is the $n$-th harmonic number. I am interested in the following "modified Euler sum": $$\sum_{n = 1}^{\infty} \frac{H_{2n}}{(2n)^3} \equiv \frac{1}{8} \sum_{n = 1}^{\infty} \frac{H_{2n}}{n^3}.$$ Can it as well be expressed in terms of $\zeta$-values? I have not been able to locate any relevant literature.

Update I: For what it may be worth (it is not even strictly related to the original question of this post, although it does feature sums containing $H_{2n}$), I have been able to derive the following identity: $$\frac{124}{5}\zeta(5) = 2\pi^{2}\ln2 + \frac{16}{\pi^{2}} \sum_{n=1}^{\infty}[\frac{6H_{n}}{(2n+1)^{4}} + \frac{H_{n}^{(2)}}{(2n+1)^{3}} - \frac{12H_{2n}}{(2n+1)^{4}} - \frac{4H_{2n}^{(2)}}{(2n+1)^{3}}],$$ where $H_{n}^{(2)} \equiv \sum_{k=1}^{n} 1/k^{2}$, as standardly. This has been done by considering a contour integral at infinity (as a limit, of course) of the function $$f(s) = \frac{\Psi(s)}{s^{3}\cos(\pi s)^{2}},$$ the corresponding sum of all residues having to vanish. However, that identity leaves me no less stuck.

Update II: Using the following family of functions: $$F_{k}(s) = \frac{\Psi(s)}{s^{k}\sin(\pi s)}[1 + \cos(\pi s)],$$ where $k$ is a natural number, I have been able to establish the following identity [note the condition on $k$]: $$\sum_{n=1}^{\infty}\frac{H_{2n}}{(2n)^{k}} = 2^{-k} \gamma \zeta(k) + 2^{-(k+2)}(k+1)\zeta(k+1) - \frac{\pi}{4} \rm{Res}(F_{k};0),\text{ for } k \text{ even},$$ where $\rm{Res}(F_{k};0)$ denotes, of course, the residue of $F(s)$ at $s = 0$. It implies the following specific cases [note the presence of $n^{k}$ in the denominators of the sums, not $(2n)^{k}$ as above] \begin{align} \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}} &= \frac{11 \zeta(3)}{4},\\ \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{4}} &= \frac{37 \zeta(5)}{4} - 4\zeta(2)\zeta(3),\\ \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{6}} &= \frac{135 \zeta(7)}{4} - 16\zeta(2)\zeta(5) - 4\zeta(4)\zeta(3). \end{align} Unfortunately, the method fails for k odd, in the sense that the resulting condition (of the sum of residues having to vanish) does not contain any harmonic numbers.

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    $\begingroup$ A CAS gives a rather simple expression except for one very ugly term. $\endgroup$ – Claude Leibovici Nov 10 '18 at 8:55
  • $\begingroup$ Thanks for your comment. Maple, which I use, seems incapable of manipulating the sum. Are you using Mathematica? $\endgroup$ – John Fredsted Nov 10 '18 at 9:03
  • $\begingroup$ No, I am using an old CAS which has been improved over the years to fit our needs in my research group. $\endgroup$ – Claude Leibovici Nov 10 '18 at 9:17
  • $\begingroup$ @ClaudeLeibovici, you have your own CAS? That's amazing $\endgroup$ – Yuriy S Nov 10 '18 at 9:18
  • $\begingroup$ @ClaudeLeibovici: Are you able to elaborate a bit on the first comment of yours? $\endgroup$ – John Fredsted Nov 10 '18 at 9:20

An attempt to get an expression through the integrals.

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\sum_{n \geq 1}\frac{1}{n^3} \int_0^1 \frac{1-x^{2n}}{1-x}dx=$$

$$\int_0^1 \frac{dx}{1-x} \left(\zeta(3)-\sum_{n \geq 1}\frac{x^{2n}}{n^3} \right)=\int_0^1 \frac{dx}{1-x} \left(\zeta(3)-\text{Li}_3 (x^2)\right)=$$

Let's change the variable to $x=e^{-t}$, then we have:

$$=\int_0^\infty \frac{ dt}{e^t-1} \left(\zeta(3)-\text{Li}_3 (e^{-2t})\right)=$$

There are known integral expressions:

$$\zeta(3)=\frac{1}{2} \int_0^\infty \frac{u^2 du}{e^u-1}$$

$$\text{Li}_3 (e^{-2t})=\frac{1}{2} \int_0^\infty \frac{u^2 du}{e^{2t+u}-1}$$

So we can write:

$$=\frac{1}{2} \int_0^\infty \int_0^\infty\frac{ dt}{e^t-1}\left( \frac{1}{e^u-1}-\frac{1}{e^{2t+u}-1} \right) u^2 du=$$

$$=\frac{1}{2} \int_0^\infty \int_0^\infty\frac{ dt}{e^t-1} \frac{e^u (e^{2t}-1)u^2 du}{(e^u-1)(e^{2t+u}-1)} = \frac{1}{2} \int_0^\infty \int_0^\infty \frac{e^u (e^t+1)u^2 du~dt }{(e^u-1)(e^{2t+u}-1)}$$

So we have:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^\infty \int_0^\infty \frac{e^u (e^t+1)u^2 du~dt }{(e^u-1)(e^{2t+u}-1)}$$

We could set:

$$e^{-t}=x \\ e^{-u}=y$$


$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \int_0^1 \frac{ (1+x) \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}$$

Which could be separated into two distinct parts:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}+\frac{1}{4} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x y)}=$$

$$=\frac{1}{2} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}-\frac{1}{4} \int_0^1 \frac{ \ln^2 y \ln(1-y) ~dy }{y(1-y)}$$

Mathematica gives for that last integral:

$$-\frac{1}{4} \int_0^1 \frac{ \ln^2 y \ln(1-y) ~dy }{y(1-y)}=\frac{\pi^4}{144}$$

So we can write (also taking the first integral w.r.t. $x$):

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \frac{ \tanh^{-1} \sqrt{y} \ln^2 y ~dy }{\sqrt{y}(1-y)}+\frac{\pi^4}{144}$$

The first integral can be simplified:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}= 4\int_0^1 \frac{ \tanh^{-1} y \ln^2 y ~dy }{1-y^2}+\frac{\pi^4}{144}$$

No idea what to do with the last integral, Mathematica can't take it. However, the integrand is nice, there's not even any singularities.

We could try integration by parts with $$u=\tanh^{-1} y \ln^2 y \\ dv=\frac{dy }{1-y^2}$$

$$I=\int_0^1 \frac{ \tanh^{-1} y \ln^2 y ~dy }{1-y^2}=(\tanh^{-1} y)^2 \ln^2 y \bigg|_0^1-I-2 \int_0^1 (\tanh^{-1} y)^2 \ln y ~\frac{dy}{y}$$


$$I=-\int_0^1 (\tanh^{-1} y)^2 \ln y ~\frac{dy}{y}= \int_0^\infty t (\tanh^{-1} e^{-t})^2 dt$$

Not really a simplification, but also an interesting expression.

Moreover, it is very fitting for numerical computation of the integral, because for moderately large $t$ we have $$\tanh^{-1} e^{-t} \asymp e^{-t}$$ and so the integral becomes elementary.

Hence, we can write:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3} = \frac{\pi^4}{144}+4\int_0^\infty t (\tanh^{-1} e^{-t})^2 dt$$

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3} \approx \frac{\pi^4}{144}+(1+2 \tau) e^{-2 \tau}+ 4\int_0^\tau t (\tanh^{-1} e^{-t})^2 dt$$

Where the last integral can be evaluated with any suitable numerical quadrature. $\tau=5$ already gives $6$ accurate digits for the series.

  • $\begingroup$ Nice way to do it ! $\to +1$ $\endgroup$ – Claude Leibovici Nov 10 '18 at 9:36
  • $\begingroup$ @ClaudeLeibovici, thank you. I made it a little less messy now $\endgroup$ – Yuriy S Nov 10 '18 at 9:39
  • $\begingroup$ Thanks a lot for your great efforts (+1 from me). However, with no intention of being ungrateful, the formulas you provide are not of the form I would like: a sum of various $\zeta$-values; or for that matter, in a closed form. An aside: I have numerically verified your last expression, now with the correct factor of 4 in front of the integral :-). $\endgroup$ – John Fredsted Nov 10 '18 at 9:57
  • $\begingroup$ @JohnFredsted, I would bet there's no expression as a finite sum of zeta values. That said, I don't claim this post answers your question, which is why I said it's an attempt. (Obviously, the term with $\pi^4$ is proportional to $\zeta(4)$) $\endgroup$ – Yuriy S Nov 10 '18 at 9:59
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    $\begingroup$ Like this? $8 Li_4(1/2) + 7 ζ(3) log(2) - π^4/15 + (log^4(2))/3 - π^{2 }log^2(2) / 3 = 1.974627536841328495420378724802799591022217356151974$831372. $\endgroup$ – Lysistrata Nov 10 '18 at 10:59

\begin{align} \sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}&=\frac12\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}\\ &=\frac12\left(\frac54\zeta(4)\right)+\frac12S\\ &=\frac58\zeta(4)+\frac12S \end{align}

\begin{align} S&=\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}=\frac12\int_0^1\frac{\ln^2x}{x}\sum_{n=1}^\infty H_n(-x)^n\ dx=-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx}_{x=(1-y)/y}\\ &=\frac12\underbrace{\int_{1/2}^1\frac{\ln^2((1-x)/x)\ln(x)}{1-x}\ dx}_{x=1-y}=\frac12\int_0^{1/2}\frac{\ln^2(x/(1-x))\ln(1-x)}{x}\ dx\\ &=\frac12\left(\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{x}\ dx+\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx\right)-\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac12\left(I_1+I_2\right)-I_3 \end{align} Applying IBP for the first integral by setting $dv=\ln x/x$ and $u=\ln(1-x)$ and letting $x=1-y$ for the second integral, we have:

\begin{align} I_1+I_2&=\frac13\ln^42+\frac13\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42+\int_0^1\frac{\ln^3x}{1-x}\ dx-\frac23\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42-6\zeta(4)-\frac23\sum_{n=1}^\infty\int_0^{1/2}x^{n-1}\ln^3x\ dx\\ &=\frac13\ln^42-6\zeta(4)+\frac23\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}\right)\\ &=4\operatorname{Li_4}\left(\frac12\right)+4\ln2\operatorname{Li_3}\left(\frac12\right)+2\ln^22\operatorname{Li_2}\left(\frac12\right)+\ln^42-6\zeta(4) \end{align} Applying IBP for the third integral by setting $dv=\ln x/x$ and $u=\ln^2(1-x)$, \begin{align} I_3=\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx&=\frac12\ln^42+\underbrace{\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\ dx}_{x=1-y}\\ &=\frac12\ln^42+\int_{1/2}^1\frac{\ln x\ln^2(1-x)}{x}\ dx \end{align} By adding the third integral to both sides, we get: \begin{align} I_3&=\frac14\ln^42+\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1 x^{n-1}\ln x\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(-\frac{1}{n^2}\right)\\ &=\frac14\ln^42+\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^3}\\ &=\frac14\ln^42-\frac14\zeta(4) \end{align} Grouping $I_1, I_2$ and $I_3$: \begin{align} S&=2\operatorname{Li_4}\left(\frac12\right)+2\ln2\operatorname{Li_3}\left(\frac12\right)+\ln^22\operatorname{Li_2}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac14\ln^42\\ &=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42 \end{align} note that we used $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$ Finally $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}=\operatorname{Li_4}\left(\frac12\right)-\frac34\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$

  • $\begingroup$ Thanks for your great efforts (point given). With the help of Maple, I have numerically 'verified' your final expression. But, very sorry to say, it is not of any use to me as it contains the polylogarithm Li(4,1/2), unless of course it can be explicitly evaluated like Li(2,1/2) and Li(3,1/2). $\endgroup$ – John Fredsted May 10 at 13:40
  • $\begingroup$ thank you. I don't think other solutions wont include $Li(4,1/2)$ . $\endgroup$ – Ali Shather May 10 at 17:58
  • $\begingroup$ I believe you are right. $\endgroup$ – John Fredsted May 11 at 5:36

Tim Jameson's notes could be helpful: Polylogarithms, multiple zeta values, and the series of Hjortnaes and Comtet, https://www.maths.lancs.ac.uk/jameson/polylog.pdf

  • $\begingroup$ Thanks for your reply. It seems to me, though, that the given article contains no $H_{2n}$, but only $H_{n}$ (and $H_{n-1}$). $\endgroup$ – John Fredsted Nov 10 '18 at 9:14
  • $\begingroup$ Apologies. I thought the biblio might point you to something useful. The only other good recent references to similar work is by Jonathan Borwein (with Zucker and Boersma) in the Ramanujan Journal, and some other notes of his scattered around Uni New South Wales websites and some he wrote with David Bailey. $\endgroup$ – Lysistrata Nov 10 '18 at 9:25
  • $\begingroup$ No need to apologize. I did not look into the references. Perhaps I should have a go at that. I will let you know if I find something suited for my needs. $\endgroup$ – John Fredsted Nov 10 '18 at 9:47
  • $\begingroup$ It's taking a while to sift through my notes and mental detritus. Your problem is almost identical to the equation at the bottom of page 12 of Elementary methods for evaluating Jordan’s sums and analogous Euler’s type sums and for setting a σ-sum theorem, G. Bastien. arxiv.org/abs/1301.7662 $\endgroup$ – Lysistrata Nov 10 '18 at 10:18
  • $\begingroup$ Lysistrata: Interesting article, getting closer. It is unfortunate, though, that $a$ is required to be an integer. If $a = 3/2$ was possible, then Eq. (20) would be spot on. The equation you allude to contains $li_{4}$. $\endgroup$ – John Fredsted Nov 10 '18 at 13:08

$$ T = \sum_{n\geq 1}\frac{H_n}{(n+1)^3}(-1)^{n+1} = \frac{1}{2}\int_{0}^{1}\frac{\log(1+x)\log^2(x)}{1+x}\,dx$$ can be seen to be equal to $$ \frac{1}{24} \left(-\pi^4-4 \pi^2 \log^2(2)+4\log^4(2)+96\,\text{Li}_4\left(\tfrac{1}{2}\right)+84\log(2)\zeta(3)\right) $$ due to the polylogarithms machinery (see De Doelder, 1991, also mentioned by Flajolet and Salvy in 1998). Of course $\sum_{n\geq 1}\frac{H_{2n}}{(2n)^3}$ is a linear combination of $T$,$\zeta(3)$ and $\zeta(4)$, hence it is inextricably related to the "ugly" terms $\log(2)\zeta(3)$ and $\text{Li}_4\left(\frac{1}{2}\right)=\sum_{n\geq 1}\frac{1}{2^n n^4}$.

  • $\begingroup$ Thanks for your reply. Why actually is $\sum_{n \ge 1} \frac{H_{2n}}{(2n)^{3}}$ a linear combination of $T$, $\zeta(3)$ and $\zeta(4)$? It is not obvious to me, but perhaps I am just being stupid. The paper by Flajolet and Salvy looks highly interesting; thanks a lot for that link. $\endgroup$ – John Fredsted Nov 11 '18 at 6:57
  • $\begingroup$ $$\sum_{n\geq 1}\frac{H_{2n}}{(2n)^3} = \frac{1}{2}\sum_{n\geq 1}\frac{H_n}{n^3}(1+(-1)^n)$$ Now $\sum_{n\geq 1}\frac{H_n}{n^3}$ is simple and the alternating sum immediately relates to the $T$ above by writing $H_n$ as $H_{n+1}-\frac{1}{n+1}$. $\endgroup$ – Jack D'Aurizio Nov 11 '18 at 16:10
  • $\begingroup$ Thanks, you are absolutely right, of course; I was indeed being stupid :-). $\endgroup$ – John Fredsted Nov 12 '18 at 9:10

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