Strenuous, arduous, laborious, onerous sum $\sum\limits_{n=1}^\infty \left(\frac1n{\sum\limits_{k=1}^n \frac1k}\right)^3$ $$\sum \limits_{n=1}^\infty \left(\frac1n{\sum \limits_{k=1}^n \frac1k}\right)^3$$
I want to see what this converges to. Obviously it converges on inspection.
Any suggestions on methods to use?
 A: $$S:=\sum_{n=1}^\infty\frac{(H_n)^3}{n^3}$$ 
is convergent since the terms are equivalent to $\,\left(\dfrac{\ln(n)}n\right)^3\,$ as $\;n\to \infty\,$ and given by $$S=\frac {31}{5040}\pi^6-\frac 52\zeta(3)^2\\\approx 2.3009545517005250398$$
I don't know a complete proof but we may start with the following class of Euler sums :
$$s_h(m,n):=\sum_{k=1}^\infty\frac{(H_k)^m}{(k+1)^n}$$
($H_k$ is the $k$-th harmonic number)
and rewrite your sum as :
\begin{align}
S&=1+\sum_{k=1}^\infty\frac{(H_{k+1})^3}{(k+1)^3}\\
&=1+\sum_{k=1}^\infty\frac{\left(H_k+\frac 1{k+1}\right)^3}{(k+1)^3}\\
&=1+s_h(3,3)+3\,s_h(2,4)+3\,s_h(1,5)+\zeta(6)-1\\
\end{align}
Using the expressions tabulated in the $1994$ paper from Bailey, Borwein and Girgensohn "Experimental Evaluation of Euler Sums" this becomes :
$$S=\left(-\frac{33}{16}\zeta(6)+2\,\zeta(3)^2\right)+3\,\left(\frac 23\zeta(6)-\frac 13\zeta(2)\zeta(4)+\frac 13\zeta(2)^3-\zeta(3)^2\right)\\+3\,\left(\frac 52\,\zeta(6)-\frac 12(2\,\zeta(4)\zeta(2)+\zeta(3)^2)\right)+\zeta(6)$$
$s_h(3,3)$ was only an experimental result when the paper initially appeared but proved since them (as indicated in the updated paper). Anyway a more direct derivation should remain of interest !
Proofs concerning $\;\displaystyle \sum_{n=1}^\infty\frac{(H_n)^2}{n^2}\;$ are provided here.
A: It converges quite quickly, so using a computer can give you the sum.  
For example in R
n <- 1:2^20
H <- cumsum(1/n)
S <- cumsum((H/n)^3)
S[2^20]

gives about 2.300955 though I suspect it is in fact just over $2.30095455$.
Alternatively use a good expansion of the Harmonic numbers, but I doubt this would be much more informative.
