$-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)=S$

Question

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 $$ \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s},\quad \zeta(2)=\frac{\pi^2}{6}. $$

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

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

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

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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.

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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*}

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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.