EDIT: Due to the solution below, I edited the answer of the post. Thanks!!!!

Hi I am trying to calculate the infinite double sum $$ S:=\sum_{j,k=1}^\infty \frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5),\quad H_n:=\sum_{k=1}^n\frac{1}{k}\ \ \ (\text{Harmonic Numbers}) $$ Thank you.

I am not sure what to do, possibly write the sum as an integral and try working with the integral instead of the sum? I was trying to figure out if we could write it as an integral representation in terms of logarithm functions. But this will just give you this sum as the answer. So I do not know how to calculate the zeta functions from the sum. Note the Riemann Zeta function is given by $$ \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s},\quad \zeta(2)=\frac{\pi^2}{6}. $$

  • $\begingroup$ where did you get this? $\endgroup$
    – cactus314
    Apr 25, 2014 at 0:00
  • $\begingroup$ This is from the early 2000's @johnmangual. It came up when I was dealing with logarithmic integrals $\endgroup$ Apr 25, 2014 at 0:02
  • $\begingroup$ Maybe write out the Harmonic sum parts and get a quadruple sum? $$ . $$Whether you can express this in terms of zeta functions or as an element of $\mathbb{Q}[\pi]$ are two separate issues. $\endgroup$
    – cactus314
    Apr 25, 2014 at 0:09
  • $\begingroup$ @johnmangual Yes this is why I have made it the bounty. It is not so simple of a problem. I apologize for not making it worth more, however I do not have any more points to give out to people. $\endgroup$ Apr 25, 2014 at 0:13
  • $\begingroup$ @user8268 Oh no, there is a nice proof of this. I know of people that have done it. It is not the easiest proof but it will be nice to see one! The trick is to use Euler sums and multiple zeta functions $\endgroup$ Apr 25, 2014 at 6:30

1 Answer 1


You can find a rigorous and short proof of this sum in a paper from Panholzer and Prodinger titled Computer-free evaluation of an infinite double sum via Euler sums dated from 2005

[Added 2014-04-27] Note: There is a typo in the paper from Panholzer and Prodinger.

The correct identity is $$S:=\sum_{j,k=1}^{\infty}\frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-{\color{red}{4}}\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)$$

Rationale: When going through the details of the proof from Panholzer and Prodinger I also did some auxiliary calculations which were not worked out in the paper. While doing so, one intermediate result in the paper, namely

\begin{align*} S&=-2\zeta(2) +\frac{1}{2}\sum_{k\geq1}\frac{H_kH_k^{(2)}}{k^2} +\frac{1}{2}\sum_{k\geq1}\frac{H_k^3}{k^2} -\frac{1}{2}\sum_{k\geq1}\frac{H_k^2}{k^3}\\ &\qquad+\left(\zeta(2)-1\right)\sum_{k\geq1}\frac{H_k}{k^2} -2\sum_{k\geq1}\frac{1}{k^2}\tag{1} \end{align*}

was somewhat peculiar to me, because the rightmost addend is $-2\zeta(2)$ and it was not clear, why it wasn't simply added to the first addend in the sum, giving $-4\zeta(2)$. My detailed calculations were conform with $(1)$. All further arguments and calculations in the paper seemed to be correct.

But, at the end the resulting value was $$S=-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)$$ Then I checked the references from the paper and I found in Carsten and Schneider the identity correctly stated with $-4\zeta(2)$. So, I came to the conclusion that there is simply a typo in the paper from Panholzer and Prodinger.

I'd like to summarize the major steps of the proof and add parts of my detailed calculations, to make the arguments above better understandable.

Task: Proof the following identity \begin{align*} S&=\sum_{j,k=1}^{\infty}\frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5) \end{align*}

The proof in the paper from Panholzer and Prodinger is done in three steps:

Step 1: Split the sum, apply partial fraction decomposition and rearrange it to get \begin{align*} S&=\sum_{j,k=1}^{\infty}\frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}\\ &=\sum_{k\geq1}\frac{H_{k+1}-1}{k(k+1)}\sum_{j\geq1}\frac{H_j}{j(j+k)}\\ &=\dots\\ &=\sum_{k\geq1}\frac{H_{k+1}-1}{k^2(k+1)}\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right)\\ \end{align*} Step 2: Apply partial fraction decomposition, index shifting, telescoping and rearrange the sum by consequently replacing all occurrences of $k+1$ with $k$. This results in \begin{align*} S&=\sum_{k\geq1}\frac{H_{k+1}-1}{k^2(k+1)}\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right)\\ &=\dots\\ &=-4\zeta(2) +\frac{1}{2}\sum_{k\geq1}\frac{H_kH_k^{(2)}}{k^2} +\frac{1}{2}\sum_{k\geq1}\frac{H_k^3}{k^2} -\frac{1}{2}\sum_{k\geq1}\frac{H_k^2}{k^3} +\left(\zeta(2)-1\right)\sum_{k\geq1}\frac{H_k}{k^2}\tag{2} \end{align*}

Note: In the paper in formula $(2)$ is the addend $-4\zeta(2)$ written as $-2\zeta(2)-2\sum_{k\geq1}\frac{1}{k^2}$

Step 3: By referring to papers from Borwein and Flajolet the following relations hold: \begin{align*} \sum_{k\geq1}\frac{H_kH_k^{(2)}}{k^2}&=\zeta(5)+\zeta(2)\zeta(3)\\ \sum_{k\geq1}\frac{H_k^3}{k^2}&=10\zeta(5)+\zeta(2)\zeta(3)\\ \sum_{k\geq1}\frac{H_k^2}{k^3}&=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)\\ \sum_{k\geq1}\frac{H_k}{k^2}&=2\zeta(3) \end{align*} Applying these relations to $(2)$ results finally in \begin{align*} S&=\sum_{j,k=1}^{\infty}\frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5) \end{align*}

And now some gory details of my calculations which I did to verify Step 2.

We consider \begin{align*} S&=\sum_{k\geq1}\frac{H_{k+1}-1}{k^2(k+1)}\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right)\\ \end{align*}

Using partial fraction decomposition the first factor of $S$ can be written as \begin{align*} \frac{H_{k+1}-1}{k^2(k+1)}&=\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_k\\ &\qquad+\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right) \end{align*} Therefore we have to evaluate:

\begin{align*} S&=\sum_{k\geq1}\left(\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_k+\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)\right)\\ &\qquad\qquad\cdot\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right) \end{align*}

We calculate it by dividing it into smaller parts. Let

\begin{align*} S_1&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_k=\sum_{k\geq1}\frac{1}{k^2}H_k-\zeta(2)\\ S_2&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)=\zeta(2)-2\\ S_3&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_k^3\\ &=\sum_{k\geq1}\frac{1}{k^2}H_k^3-3\sum_{k\geq1}\frac{1}{k^2}H_k^2+3\sum_{k\geq1}\frac{1}{k^3}H_k-\zeta(4)\\ S_4&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)H_k^2\\ &=\sum_{k\geq1}\frac{1}{k^2}H_k^2-2\sum_{k\geq1}\left(\frac{1}{k^2}+\frac{1}{k^3}\right)H_k+\zeta(3)+\zeta(4) \end{align*} \begin{align*} S_5&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_kH_k^{(2)}\\ &=\sum_{k\geq1}\frac{1}{k^2}H_kH_k^{(2)}-\sum_{k\geq1}\frac{1}{k^3}H_k-\sum_{k\geq1}\frac{1}{k^2}H_k^{(2)}+\zeta(4)\\ S_6&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)H_k^{(2)}\\ &=\sum_{k\geq1}\frac{1}{k^2}H_k^{(2)}-\zeta(3)-\zeta(4)\\ S_7&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)\frac{H_k^2}{k}\\ &=\sum_{k\geq1}\left(-\frac{1}{k^2}+\frac{1}{k^3}\right)H_k^2 +2\sum_{k\geq1}\frac{1}{k^2}H_{k}-\zeta(3)\\ S_8&=\sum_{k\geq1}\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)\frac{H_k}{k}\\ &=-2\sum_{k\geq1}\frac{1}{k^2}H_k+2\zeta(2)+\zeta(3) \end{align*}

This results in: \begin{align*} S&=\sum_{k\geq1}\frac{H_{k+1}-1}{k^2(k+1)}\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right)\\ &=\left(\left(-\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k+1}\right)H_k+\left(-\frac{1}{k}+\frac{1}{k+1}+\frac{1}{(k+1)^2}\right)\right)\\ &\qquad\cdot\left(\zeta(2)+\frac{H_k^2+H_k^{(2)}}{2}-\frac{H_k}{k}\right)\\ &=\zeta(2)(S_1+S_2)+\frac{1}{2}(S_3+S_4+S_5+S_6)-S_7-S_8\\ &=-4\zeta(2)+\frac{1}{2}\sum_{k\geq1}\frac{1}{k^2}H_kH_k^{(2)} +\frac{1}{2}\sum_{k\geq1}\frac{1}{k^2}H_k^3 -\frac12\sum_{k\geq1}\frac{1}{k^3}H_k^2 +\left(\zeta(2)-1\right)\sum_{k\geq1}\frac{1}{k^2}H_k \end{align*} which was my verification for Step 2.

  • 1
    $\begingroup$ This is quite impressive and is exactly what I was looking for. I will close this bounty now. You are well deserved of this answer. +250 :) Thank you very much. I will be back to ask you questions in regards to all these details. $\endgroup$ Apr 27, 2014 at 16:12
  • $\begingroup$ Thanks and you're welcome! Regards. $\endgroup$ Apr 27, 2014 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .