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