Is there a way to find out a closed form of the sum $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^6}\sum_{k=1}^n \dfrac{H_k}{k}$ ?

It appeared to me while calculating $\displaystyle\sum_{n=1}^\infty \dfrac{(H_n)^2}{n^6}$ where I used $\displaystyle (H_n)^2=2\sum_{k=1}^n \dfrac{H_k}{k}-H_n^{(2)}$ .

The sum involving harmonic number of order 2 can be calculated, but how do we do the mentioned sum , I still do not have much ideas about that.


Through a sequence of interchanging the order of the sums & removing the first term ... We will manipulate the sum to multiple zeta functions. \begin{eqnarray*} S &=& \sum_{n=1}^{\infty} \frac{1}{n^6} \sum_{k=1}^{n} \frac{H_k}{k} \\ S &=& \sum_{n=1}^{\infty} \sum_{k=1}^{n} \sum_{j=1}^{k}\frac{1}{n^6 jk} \\ S &=& \sum_{k=1}^{\infty} \sum_{j=1}^{k} \sum_{n=k}^{\infty}\frac{1}{n^6 jk} \\ S &=& \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} \sum_{n=k}^{\infty}\frac{1}{n^6 jk} \\ S &=& \sum_{j=1}^{\infty} \left( \sum_{n=k}^{\infty}\frac{1}{n^6 j^2}+\sum_{k=j+1}^{\infty} \sum_{n=k}^{\infty}\frac{1}{n^6 jk} \right)\\ S &=& \sum_{j=1}^{\infty} \left(\frac{1}{j^8} +\sum_{n=k+1}^{\infty}\frac{1}{n^6 j^2}+\sum_{k=j+1}^{\infty} \left( \frac{1}{k^7 j} + \sum_{n=k+1}^{\infty}\frac{1}{n^6 jk} \right)\right)\\ \end{eqnarray*} So the sum is $\color{red}{\zeta(8)+\zeta(6,2)+\zeta(7,1)+\zeta(6,1,1)}$.

These multiple zeta functions can be evaluated using techniques descibed ... https://en.wikipedia.org/wiki/Multiple_zeta_function

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    $\begingroup$ \begin{equation} S= \zeta(6,2)+115/24 \zeta(8)- 4 \zeta(3) \zeta(5)+1/2 \zeta(3)^2 \zeta(2) \end{equation} where we used the math.stackexchange.com/questions/2323993/… and the links therein. It is not known whether $\zeta(6,2)$ is reduced to single zeta values. $\endgroup$ – Przemo Jun 29 '17 at 12:04
  • $\begingroup$ @Przemo Amusingly enough I have been recently looking into these calculations to express multivariate zeta functions in terms of pure zeta values. I have got as far as length $3$, weight $5$ at the moment. It is my intention to complete the above calculation, but I suspect it will take me some time to get as far as length $3$, weight $8$. $\endgroup$ – Donald Splutterwit Jun 29 '17 at 19:15

Partial solution and just writing the integral representation of the sum.

\begin{align} S=\sum_{n=1}^\infty\frac1{n^6}\sum_{k=1}^n\frac{H_k}{k}=\frac12\sum_{n=1}^\infty\frac{H_n^2+H_n^{(2)}}{n^6} \end{align} From here we have $$\int_0^1 x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$ Thus $$S=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n^5}\ dx=\frac12\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_5(x)}{x}\ dx$$

Another related integral:

@SuperAboud proved here : $$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n(H_n^2-H_n^{(2)})$$

Multiply both sides by $\displaystyle-\frac{\ln^5x}{5!x}$ then integrate from $x=0$ to $x=1$

$$-\frac1{5!}\sum_{n=1}^\infty (H_n^2-H_n^{(2)})\int_0^1 x^{n-1}\ln^5x\ dx=\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^6}=-\frac1{5!}\int_0^1\frac{\ln^2(1-x)\ln^5x}{x(1-x)}\ dx$$

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