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I proved the following result

$$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$

After consideration of powers of polylogarithms.

You can refer to the following thread .

My question is : are there any papers in the literature which dealt with that result?

Are my evaluations worth publishing ?

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  • $\begingroup$ What does $H_k^{(2)}$ represent? A hyperharmonic number? $\endgroup$ – David Zhang Apr 28 '14 at 13:07
  • $\begingroup$ @DavidZhang, $$H_k^{(2)}=\sum_{n=1}^k\frac{1}{n^2}$$ $\endgroup$ – Zaid Alyafeai Apr 28 '14 at 13:47
  • $\begingroup$ This one has a lot of information. $\endgroup$ – Felix Marin Jun 21 '14 at 20:28
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Borwein and Girgensohn's paper "Evaluation of Triple Euler Sums" (Electronic Journal of Combinatorics 3(1) 1996) deals exactly with results of this kind. They index their problems slightly differently, so there's a little more work to be done to obtain your result, but their results are close enough to yours that I wouldn't really consider your result new. Absent newness, you might have a shot at something publishable if your proof technique is sufficiently novel or interesting. My sense, though, is that polylogarithms are one of the main proof techniques for evaluating Euler sums. For example, searching for "polylog" on Hoffman's "References on Multiple Zeta Values and Euler Sums" page generates 41 hits. So my guess is that the proof technique would not be sufficiently novel to justify publication, either.

I'll outline the steps that can be used to obtain your result from those in Borwein and Girgensohn's paper.

First, rewrite your sum as $$ \begin{align} \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} &= \sum_{k\geq 1} \frac{H_{k-1}^{(2)} H_{k-1}}{k^3} + \sum_{k\geq 1} \frac{H_{k-1}^{(2)}}{k^4} + \sum_{k\geq 1} \frac{ H_{k-1}}{k^5} +\sum_{k\geq 1} \frac{1}{k^6}\\ &= \sum_{k\geq 1} \frac{H_{k-1}^{(2)} H_{k-1}}{k^3} + \sum_{k\geq 1} \frac{H_{k}^{(2)}}{k^4} + \sum_{k\geq 1} \frac{ H_{k}}{k^5} - \sum_{k\geq 1} \frac{1}{k^6}.\\ \end{align} $$ The second sum on the previous line is $\zeta(3)^2 - \frac{1}{3}\zeta(6)$, the third is $\frac{7}{2} \zeta(6) - \zeta(4)\zeta(2) - \frac{1}{2} \zeta(3)^2$, and the last is of course $\zeta(6)$. (The second and third sums are both due to Euler. For references, see the table on page 16 and Theorem 2.2 in Flajolet and Salvy's "Euler Sums and Contour Integral Representations," Experimental Mathematics 7 1998, pp. 15-35.)

The first sum is the kind that Borwein and Girgensohn show how to evaluate. Theorem 2 of their paper says that $$\sum_{k =1}^N \frac{H_{k-1}^{(2)} H_{k-1}}{k^3} = \zeta_N(3,2,1) + \zeta_N(3,1,2) + \zeta_N(3,3),$$ where $$ \begin{align} \zeta_N(a,b) &= \sum_{i=1}^N \sum_{j=1}^{i-1} \frac{1}{i^a j^b}, \\ \zeta_N(a,b,c) &= \sum_{i=1}^N \sum_{j=1}^{i-1} \sum_{k=1}^{j-1} \frac{1}{i^a j^b k^c}. \\ \end{align} $$ Then, on page 21, they give the evaluations, where $\zeta(a,b,c) = \lim_{N \to \infty} \zeta_N(a,b,c)$, $$ \begin{align} \zeta(3,2,1) &= 3 \zeta(3)^2 - \frac{203}{48} \zeta(6),\\ \zeta(3,1,2) &= \frac{53}{24} \zeta(6) - \frac{3}{2} \zeta(3)^2. \end{align} $$ Values of $\zeta_N(a,a)$ are classical (see, for example, Concrete Mathematics, p. 37, Eq. 2.33 for evaluations of this type). We have $$\zeta_N(3,3) = \frac{1}{2} \left( \left(\sum_{k=1}^N \frac{1}{k^3} \right)^2 - \sum_{k=1}^N \frac{1}{k^6}\right),$$ so that $$\zeta(3,3) = \lim_{N \to \infty} \zeta_N(3,3) = \frac{1}{2} \zeta(3)^2 - \frac{1}{2} \zeta(6).$$

We also need that $$ \begin{align} \zeta(4) \zeta(2) &= \frac{7}{4} \zeta(6), \\ \zeta(2)^3 &= \frac{35}{8} \zeta(6), \end{align} $$ which just follow from the known values of $\zeta(2), \zeta(4)$, and $\zeta(6)$.

Putting all of this together yields $$ \begin{align} \sum_{k=1}^{\infty} \frac{H_k^{(2)} H_k}{k^3} = &3 \zeta(3)^2 - \frac{203}{48} \zeta(6) + \frac{53}{24} \zeta(6) - \frac{3}{2} \zeta(3)^2 + \frac{1}{2} \zeta(3)^2 - \frac{1}{2} \zeta(6) + \zeta(3)^2 \\ &- \frac{1}{3}\zeta(6) + \frac{7}{2} \zeta(6) - \zeta(4)\zeta(2) - \frac{1}{2} \zeta(3)^2 - \zeta(6) \\ = &\frac{5}{2} \zeta(3)^2 - \frac{101}{48} \zeta(6), \end{align} $$ which is exactly what you have when you express $\zeta(4)\zeta(2)$ and $\zeta(2)^3$ in terms of $\zeta(6)$.

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  • $\begingroup$ Hey Mike thanks for the replay. If we wrote the Euler sum as $$\sum_{k\geq 1}\frac{H^{(a)}_k H^{(b)}_k}{k^c}$$ Then for each values of $a,b$ and $c$ is the sum reducible to zeta values ? $\endgroup$ – Zaid Alyafeai Jan 1 '14 at 12:20
  • $\begingroup$ The Borwein and Girgensohn paper proves that the answer to your question is "Yes" for certain values of $a$, $b$, and $c$. That paper is nearly 20 years old, though, and the state of the art may have improved since then. $\endgroup$ – Mike Spivey Jan 2 '14 at 18:16
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    $\begingroup$ +1. Very good. Thanks for given many interesting references. $\endgroup$ – Felix Marin Jun 21 '14 at 20:30
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An evaluation using contour integration is also possible.

Consider the function $$ f(z) = \frac{\pi \cot (\pi z) [ \psi(-z)+\gamma] \ \psi_{1}(-z)}{z^3}, $$ where $\psi(z)$ is the digamma function, $\gamma$ is the Euler-Mascheroni constant, and $\psi_{1}(z)$ is the trigamma function.

The function $f(z)$ has poles at the positive integers of order 4, simple poles at the negative integers, and a pole of order 7 at the origin.

On the sides of a square with vertices at $z= (N+\frac{1}{2})\pm i (N+ \frac{1}{2})$ (call it $C_{N}$), $\cot (\pi z)$ is uniformly bounded.

And when $z$ is large in magnitude and not on the positive real axis, $\psi(-z) \sim \log(-z)$ and $\psi_{1}(-z) \sim - \frac{1}{z}$.

So $ \displaystyle \int_{C_{N}} f(z) \ dz \to 0$ as $N \to \infty$ through the positive integers, and therefore

$$ \sum_{n=-\infty}^{\infty} \text{Res}[f(z), n] = 0.$$

Expanding at the positive integers, we get

$$ \begin{align} f(z) &= \frac{1}{z^{3}} \Bigg[ \Bigg(\frac{1}{z-n} - 2 \zeta(2) (z-n) + \mathcal{O} \Big((z-n)^{3} \Big) \Bigg) \Bigg(\frac{1}{z-n} + H_{n} - \Big(H_{n}^{(2)}+\zeta(2) \Big)(z-n)\\ &+ \Big(H_{n}^{(3)} - \zeta(3) \Big) \cdot(z-n)^{2}+ \mathcal{O} \big((z-n)^{3} \Bigg) \Bigg( \frac{1}{(z-n)^{2}} + \Big( H_{n}^{2} + \zeta(2) \Big) - 2 \Big(H_{n}^{3} - \zeta(3) \Big) \\ & \cdot (z-n) + \mathcal{O} \Big( (z-n)^{2} \Big) \Bigg) \Bigg] \\ &= \frac{1}{z^{3}} \left(\frac{1}{(z-n)^{4}} + \frac{H_{n}}{(z-n)^{2}} - \frac{2 \zeta(2)}{(z-n)^{2}} + \frac{H_{n}^{(2)} H_{n}}{z-n} - \frac{\zeta(2) H_{n}}{z-n} + \frac{\zeta(3)}{z-n} - \frac{H_{n}^{(3)}}{z-n} + \mathcal{O}(1) \right) .\end{align}$$

So at the positive integers, $$ \begin{align} \text{Res}[f(z),n] &= \text{Res} \left[ \frac{1}{z^{3}} \frac{1}{(z-n)^{4}}, n \right] + \text{Res} \left[ \frac{1}{z^{3}} \frac{H_{n}}{(z-n)^{3}}, n \right] + \text{Res} \left[ \frac{-2 \zeta(2)}{z^{3}} \frac{1}{(z-n)^{2}}, n \right] \\ &+ \text{Res} \Big[ \frac{H_{n}^{(2)} H_{n}}{z^{3}} \frac{1}{z-n}, n \Big] + \text{Res} \Big[ \frac{- \zeta(2) H_{n}}{z^{3}} \frac{1}{z-n}, n \Big] + \text{Res} \Big[ \frac{\zeta(3)}{z^{3}} \frac{1}{z-n}, n \Big] \\ &+ \text{Res} \Big[ \frac{-H_{n}^{(3)}}{z^{3}} \frac{1}{z-n}\Big] \\ &= - \frac{10}{n^{5}} + \frac{6 H_{n}}{n^{5}} + \frac{6 \zeta(2)}{n^{4}} + \frac{H_{n}^{(2)}H_{n}}{n^{3}} - \frac{\zeta(2) H_{n}}{n^{3}} + \frac{\zeta(3)}{n^{3}} - \frac{H_{n}^{(3)}}{n^{3}} .\end{align} $$

At the negative integers, $$\begin{align} \text{Res}[f(z),-n] &= -\frac{\psi_{1}(n) \Big( \psi(n)+\gamma\Big)}{n^3} \\ &= - \frac{\Big(\zeta(2) - H_{n-1}^{(2)} \Big)H_{n-1}}{n^{3}} \\ &= -\frac{ \Big(\zeta(2)-H_{n}^{(2)}+ \frac{1}{n^{2}}\Big) \Big(H_{n}- \frac{1}{n} \Big)}{n^{3}} \\ &= - \frac{\zeta(2) H_{n}}{n^{3}} + \frac{\zeta(2)}{n^{4}} + \frac{H_{n}^{(2)} H_{n}}{n^{3}} - \frac{H_{n}^{(2)}}{n^{4}} - \frac{H_{n}}{n^{5}} + \frac{1}{n^{6}}. \end{align}$$

And expanding at the origin, we get $$ \begin{align} f(z) &= \frac{1}{z^{3}} \Big( \frac{1}{z} - 2 \zeta(2) z - 2 \zeta(4) z^{3} - 2 \zeta(6) z^{5} + \mathcal{O}(z^{7}) \Big) \Big( \frac{1}{z} - \zeta(2)z - \zeta(3) z^{2} - \zeta(4) z^{3}- \zeta(5) z^{4} \\ &- \zeta(6) z^{5} + \mathcal{O}(z^{6}) \Big) \Big( \frac{1}{z^{2}} + \zeta(2) + 2 \zeta(3) z + 3 \zeta(4) z^{2} + 4 \zeta(5)z^{3} + 5 \zeta(6)z^{4} + \mathcal{O}(z^{5}) \Big) \\ &= \frac{1}{z^{7}} - \frac{2 \zeta(2)}{z^{5}} + \frac{\zeta(3)}{2z^{4}} - \frac{\zeta^{2}(2)}{z^{3}} + \frac{3 \zeta(5)}{z^{2}} - \frac{5 \zeta(2) \zeta(3)}{z^{2}} - \frac{2 \zeta^{2}(3)}{z} + \frac{2 \zeta(6)}{z} - \frac{8 \zeta(2) \zeta(4)}{z} \\ &+ \frac{2 \zeta^{3}(2)}{z} + \mathcal{O}(1) .\end{align}$$

Therefore, $$ \text{Res}[f(z),0] = -2 \zeta^{2}(3) + 2 \zeta(6) - 8 \zeta(2) \zeta(4) + 2 \zeta^{3}(2) .$$

Summing up all the residues,

$$ \begin{align} &-10 \sum_{n=1}^{\infty} \frac{1}{n^{6}} + 6 \sum_{n=1}^{\infty} \frac{H_{n}}{n^{5}} + 6 \zeta(2) \sum_{n=1}^{\infty} \frac{1}{n^{4}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)} H_{n}}{n^{3}} - \zeta(2) \sum_{n=1}^{\infty} \frac{H_{n}}{n^{3}} + \zeta(3) \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\ &- \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}} - \zeta(2) \sum_{n=1}^{\infty} \frac{H_{n}}{n^{3}} + \zeta(2) \sum_{n=1}^{\infty} \frac{1}{n^{4}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)} H_{n}}{n^{3}} - \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} - \sum_{n=1}^{\infty} \frac{H_{n}}{n^{5}} + \sum_{n=1}^{\infty} \frac{1}{n^{6}} \\ &- 2 \zeta^2(3) + 2 \zeta(6) - 8 \zeta(2) \zeta(4) + 2 \zeta^3(2) = 0, \end{align} $$

where $$ \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k) \tag{1}$$

and $$\sum_{n=1}^{\infty} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}. \tag{2}$$

Therefore, $$ 2 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}H_{n}}{n^{3}} = -10 \zeta(6) + \frac{17}{2} \zeta(2) \zeta(4) + 4 \zeta^{2}(3) - 2 \zeta^{3}(2) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} .$$

To evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}}$ consider $$ g(z) = \frac{\Big(\psi_{1}(-z) \Big)^{2}}{z^{\color{red}{3}}} $$ and integrate around the same contour.

The function $g(z)$ has poles of order 4 at the positive integers and a pole of order 7 at the origin.

Then again because the integral vanishes as $ N \to \infty$ through the positive integers, $$\sum_{n=0}^{\infty} \text{Res}[g(z), n] =0.$$

Expanding at the positive integers, we get $$ \begin{align} g(z) &= \frac{1}{z^{3}} \Bigg[\frac{1}{(z-n)^{2}}+ \Big(H_{n}^{(2)} + \zeta(2) \Big) - 2 \Big(H_{n}^{(3)}-\zeta(3)\Big) (z-n) + \mathcal{O}\Big((z-n)^{2}\Big) \Bigg]^{2} \\ &= \frac{1}{z^{3}} \Bigg(\frac{1}{(z-n)^{4}} + \frac{2 H_{n}^{(2)}+2 \zeta(2)}{(z-n)^{2}} - \frac{4H_{n}^{(3)} -4 \zeta(3)}{z-n} + \mathcal{O}(1) \Bigg) .\end{align}$$

So at the positive integers, $$ \begin{align} \text{Res} [g(z),n] &= \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{1}{(z-n)^{4}},n \Bigg] + \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{2 H_{n}^{(2)} + 2 \zeta(2)}{(z-n)^{2}},n\Bigg] + \text{Res} \Bigg[ \frac{4 \zeta(3) - 4 H_{n}^{(3)}}{z-n},n\Bigg] \\ &= -\frac{10}{n^{6}} - \frac{6 H_{n}^{(2)}}{n^{4}} - \frac{6 \zeta(2)}{n^{4}} - \frac{4 H_{n}^{(3)}}{n^{3}} + \frac{4 \zeta(3)}{n^{3}} . \end{align}$$

And expanding at the origin, we get $$ \begin{align} g(z) &= \frac{1}{z^{3}} \Big( \frac{1}{z^{2}} + \zeta(2) + 2 \zeta(3) z + 3 \zeta(4) z^{2} + 4 \zeta(5) z^{3} + 5 \zeta(6)z^{4} + \mathcal{O}(z^{5}) \Big)^{2} \\ &= \frac{1}{n^{7}} + \frac{2 \zeta(3)}{z^{5}} + \frac{2 \zeta(3)}{z^{4}} + \frac{6 \zeta(4) + \zeta^{2}(z)}{z^{3}} + \frac{9 \zeta(5) + 4 \zeta(2) \zeta(3)}{z^{2}} \\ &+ \frac{10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3)}{z} + \mathcal{O}(1) . \end{align}$$

Therefore, $$ \text{Res}[g(z),0] = 10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3) .$$

Summing up all the residues, $$\displaystyle -10 \zeta(6) - 6 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} - 6 \zeta(2) \zeta(4) - 4 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}} + 4 \zeta^{2}(3) + 10 \zeta(6) + 6 \zeta(2) \zeta(4) + 4 \zeta^{2}(3) =0, $$

which implies $$ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} = \zeta^{2}(3) - \frac{\zeta(6)}{3} .$$

Combining this result with the previous result, we get $$ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}H_{n}}{n^{3}} = - \frac{31}{6} \zeta(6) + \frac{17}{4} \zeta(2) \zeta(4) + \frac{5}{2} \zeta^{2}(3) - \zeta^{3}(2) \approx 1.4716926365,$$

which is equivalent to your answer.

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$(1)$ Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

$(2)$ Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$

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Using the following identity (see here): $$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac1n\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right)$$ divide both sides by $n^2$ then take the sum \begin{align} I&=\color{blue}{-\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}}\\ &=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n^2}\ dx=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\\ &=\int_0^1\frac{\ln^3x\operatorname{Li}_2(1-x)}{1-x}\ dx=\int_0^1\frac{\ln^3(x)}{1-x}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\ln^3x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=\zeta(2)(-6\zeta(4))+\sum_{n=1}^\infty\left(H_n-\frac1n\right)\int_0^1x^{n-1}\ln^4x\ dx-\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^3x\ dx\\ &=-\frac{21}2\zeta(6)+\sum_{n=1}^\infty\left(H_n-\frac1n\right)\left(\frac{24}{n^5}\right)-\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\left(-\frac{6}{n^4}\right)\\ &=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{n^5}-24\zeta(6)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-6\zeta(6)\\ &=-\frac{81}2\zeta(6)+24\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}\\ &=\color{blue}{\frac{3}2\zeta(6)-12\zeta^2(3)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}} \end{align}

Rearranging the blue terms, we get \begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}=4\zeta^2(3)-\frac12\zeta(6)-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-\frac23\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}-\frac13\sum_{n=1}^\infty\frac{H_n^3}{n^3}\tag{1} \end{align} Lets simplify more and starting with the following sum \begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=\sum_{n=1}^\infty \frac1{n^3}\sum_{k=1}^\infty\left(\frac{1}{k^3}-\frac{1}{(k+n)^3}\right)=\zeta^2(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{1}{n^3(n+k)^3}\right)\\ &=\zeta^2(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\left(\frac{6}{k^5}\left(\frac1n-\frac1{n+k}\right)-\frac{3}{k^4n^2}-\frac{3}{k^4(k+n)^2}+\frac{1}{k^3n^3}-\frac{1}{k^3(k+n)^4}\right)\right)\\ &=\zeta^2(3)-\sum_{k=1}^\infty \left(\frac{6H_k}{k^5}-\frac{3\zeta(2)}{k^4}-\frac{3}{k^4}\left(\zeta(2)-H_k^{(2)}\right)+\frac{\zeta(3)}{k^3}-\frac{1}{k^3}\left(\zeta(3)-H_k^{(3)}\right)\right)\\ &=\zeta^2(3)-6\sum_{k=1}^\infty\frac{H_k}{k^5}+3\zeta(2)\zeta(4)+3\zeta(2)\zeta(4)-3\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}-\zeta^2(3)+\zeta^2(3)-S\\ 2S&=\zeta^2(3)-6\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)+\frac{21}{2}\zeta(6)-3\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4} \end{align} which follows \begin{align} \sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\tag{2} \end{align} Plugging $(2)$ in $(1)$ we have \begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}=\frac43\zeta^2(3)-\frac12\zeta(6)+\frac23\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}-\frac13\sum_{n=1}^\infty\frac{H_n^3}{n^3} \end{align} Using the formula $\ \displaystyle \sum_{n=1}^\infty\frac{H_n^{(a)}}{n^a}=\frac12(\zeta(2a)+\zeta^2(a))\ $ gives us $\ \displaystyle \sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=\frac12(\zeta(6)+\zeta^2(3))$

Cornel Ioan Valean was able here to prove $\ \displaystyle \sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$

Substituting these two sums, finally we get

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}=\frac52\zeta^2(3)-\frac{101}{48}\zeta(6)$$

And as a bonus, in $(2)$, we got $$ \sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}=\zeta^2(3)-\frac13\zeta(6)$$

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The following new solution is proposed by Cornel Ioan Valean. Based on a few ideas presented in the book, (Almost) Impossible Integrals, Sums, and Series, like the Cauchy product of $(\operatorname{Li}_2(x))^2$, that is $\displaystyle (\operatorname{Li}_2(x))^2=4\sum_{n=1}^{\infty}x^n\frac{H_n}{n^3}+2\sum_{n=1}^{\infty}x^n\frac{H_n^{(2)}}{n^2}-6\sum_{n=1}^{\infty}\frac{x^n}{n^4}$, where if we multiply both sides by $\displaystyle \frac{\log(1-x)}{x}$ and then integrate from $x=0$ to $x=1$, using that $\displaystyle \int_{0}^{1}x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_{n}}{n}$, we get \begin{equation*} \int_0^1 \frac{\log(1-x)}{x}(\operatorname{Li}_2(x))^2 \textrm{d}x=-\frac{1}{3}(\operatorname{Li}_2(x))^3\biggr|_{x=0}^{x=1}=-\frac{35}{24}\zeta(6) \end{equation*} \begin{equation*} =6\sum_{n=1}^{\infty} \frac{H_n}{n^5}-4\sum_{n=1}^{\infty} \frac{H_n^2}{n^4}-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3} \end{equation*} \begin{equation*} =5\zeta^2(3)-\frac{17}{3}\zeta(6)-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3}, \end{equation*} where the first sum comes from the classical generalization, $ \displaystyle 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\in \mathbb{N},\ n\ge2$, and the second sum, $\displaystyle \sum_{n=1}^{\infty} \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$, is calculated in the mentioned book or in this article.
To conclude, we have \begin{equation*} \sum_{n=1}^{\infty}\frac{H_n H_n^{(2)}}{n^3}=\frac{1}{2}\left(5\zeta^2(3)-\frac{101}{24}\zeta(6)\right). \end{equation*} Note the present solution circumvents the necessity of using the value of the series $\displaystyle \sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3$.

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    $\begingroup$ yes i saw this answer on his FB page. really beautiful solution specially when we didnt need the value of the cubic sum. this solution is considered the shortest solution to find the value of the cubic sum as we can relate the two sums as explained in (1) . $\endgroup$ – Ali Shather Jun 30 at 9:48
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Second approach with a big bonus: We are going to establish two relations and solve them by elimination.

The first relation: Using the following identity (see here): $$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac1n\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right)$$ divide both sides by $n^2$ then take the sum \begin{align} R_1&=\color{blue}{-\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}}\\ &=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n^2}\ dx=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\\ &=\int_0^1\frac{\ln^3x\operatorname{Li}_2(1-x)}{1-x}\ dx=\int_0^1\frac{\ln^3(x)}{1-x}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\ln^3x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=\zeta(2)(-6\zeta(4))+\sum_{n=1}^\infty\left(H_n-\frac1n\right)\int_0^1x^{n-1}\ln^4x\ dx-\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^3x\ dx\\ &=-\frac{21}2\zeta(6)+\sum_{n=1}^\infty\left(H_n-\frac1n\right)\left(\frac{24}{n^5}\right)-\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\left(-\frac{6}{n^4}\right)\\ &=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{n^5}-24\zeta(6)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-6\zeta(6)\\ &=-\frac{81}2\zeta(6)+24\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}\\ &=\color{blue}{\frac{3}2\zeta(6)-12\zeta^2(3)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}} \end{align}

Thus
\begin{align} \boxed{R_1=\sum_{n=1}^\infty\frac{H_n^3}{n^3}+3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=12\zeta^2(3)-\frac32\zeta(6)-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}} \end{align}


The second relation: From this solution, we have $$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Multiply both isdes by $\frac{\ln^2x}{2x}$ then integrate from $x=0$ to $1$, we get

\begin{align} R_2&=\frac12\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln^2x\ dx\\ &=\color{blue}{\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}}\\ &=-\frac12\int_0^1\frac{\ln^3(1-x)\ln^2x}{x(1-x)}\ dx\overset{\color{red}{1-x\ \mapsto\ x}}{=}-\frac12\int_0^1\frac{\ln^3x\ln^2(1-x)}{x(1-x)}\ dx\tag{i}\\ &=-\frac12\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ln^3x\ dx\color{blue}{=3\sum_{n=1}^\infty\frac{H_n^2}{n^4}-3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}} \end{align}

Thus
\begin{align} \boxed{R_2=\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=3\sum_{n=1}^\infty\frac{H_n^2}{n^4}-3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}}\\ \end{align} Note that in $(i)$ we used $\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n(H_n^2-H_n^{(2)}).$


Therefore: $$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{R_1+R_2}{2}=6\zeta^2(3)-\frac34\zeta(6)-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}+3\sum_{n=1}^\infty\frac{H_n^2}{n^4}-9\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}$$

and $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}=\frac{R_1-R_2}{6}=2\zeta^2(3)-\frac14\zeta(6)-\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^4}-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}$$

Substituting the following results: $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=\frac12\zeta(6)+\frac12\zeta^2(3)$$ $$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$ $$\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)$$

Finally we get: $$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$$ and $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^3}=\frac52\zeta^2(3)-\frac{101}{48}\zeta(6)$$

Note: $\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}$ are proved in my first solution above. As for $\sum_{n=1}^\infty\frac{H_n^2}{n^4}$, can be found here.

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