Is there a way to find out a closed form of the sum $\sum_{n=1}^\infty \frac{1}{n^6}\sum_{k=1}^n \frac{H_k}{k}$? 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.
 A: 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
A: 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 $x=1$
$$-\frac1{5!}\sum_{n=1}^\infty (H_n^2-H_n^{(2)})\int_0^1 x^{n-1}\ln^5x\ dx=\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^6}=-\frac1{5!}\int_0^1\frac{\ln^2(1-x)\ln^5x}{x(1-x)}\ dx$$
