Beautiful Double Sum

Consider the sum : $\displaystyle S=\sum_{n=1}^{\infty} {\rm H}_n (\frac{1}{n}-\frac{1}{n+k})$

Note that , $\displaystyle S=\sum_{m=1}^{n}\frac{1}{m} \sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+k}) = \sum_{m\ge 1}\frac{1}{m}\sum_{n=m}^{\infty} {\rm H}_n (\frac{1}{n}-\frac{1}{n+k})$

By telescoping & partial fraction decomposition we have , $\displaystyle S= \zeta(2) + \sum_{n=1}^{k-1} \frac{{\rm H}_n}{n} = \zeta(2) + \frac{1}{2}({\rm H}_{k-1}^{(2)} + ({\rm H}_{k-1})^2)$

The main sum $\displaystyle I = \sum _{ k=1 }^{ \infty }{ \sum _{ j=1 }^{ \infty }{ \left[ \frac { { H }_{ j }\left( { H }_{ k+1 }-1 \right) }{ kj\left( k+1 \right) \left( j+k \right) } \right] } } = \sum_{k\ge 1} \frac{{\rm H}_{k+1}-1}{k^2(k+1)} \sum_{j\ge 1} \frac{{\rm H}_j}{j(j+k)}$

Using the above result and expanding the main sum after substituting values for the inner sum we have ,

$\displaystyle I = \frac{1}{2} \sum_{n\ge 1}\frac{{\rm H}_n ({\rm H}_{n})^2}{n^2} + \frac{1}{2}\sum_{n\ge 1} \frac{{\rm H}_{n}^{3}}{n^2}-\sum_{n\ge 1} \frac{{\rm H}_{n}^{2}}{n^3}+\frac{\pi^2 -6 }{6}\sum_{n\ge 1}\frac{{\rm H}_{n}}{n^2} - \zeta(2)$

Categorising our four sums i.e. ,

$\displaystyle \sum_{n\ge 1}\frac{{\rm H}_n ({\rm H}_{n})^{(2)}}{n^2} \text{ (*)}$

$\displaystyle \sum_{n\ge 1} \frac{{\rm H}_{n}^{3}}{n^2} \text{ (**)}$

$\displaystyle \sum_{n\ge 1} \frac{{\rm H}_{n}^{2}}{n^3} \text{ (***)}$

$\displaystyle \sum_{n\ge 1}\frac{{\rm H}_{n}}{n^2} \text{ (****)}$

Before we get into the sums define the multiple zeta functions as , $\displaystyle \zeta(a,b,c) = \sum_{m=1}^{\infty} \sum_{n=1}^{m-1}\sum_{p=1}^{n-1} \frac{1}{m^a n^b p^c}$ & $\displaystyle \zeta(a,b) = \sum_{m=1}^{\infty} \sum_{n=1}^{m-1} \frac{1}{m^a n^b}$

$\text{****}$ Easily follows from euler sums and is $\displaystyle \sum_{n\ge 1} \frac{{\rm H}_n}{n^2} = 2\zeta(3)$

For $\text{***}$ , Let $\displaystyle J = \sum_{n\ge 1} \frac{{\rm H}_{n}^{2}}{n^3}$

A relationship between MZV & MHS values states that $\displaystyle \sum_{n=1}^{\infty} \frac{{\rm H}_n(a_1,a_2,a_3,\cdots,a_k)}{n^a} = \zeta(a_1,a_2,\cdots,a_k)+\zeta(a+a_1,a_2,a_3,\cdots,a_k)$

The quasi-shuffle identity shows that , $\displaystyle {\rm H}_n^2(1) = 2{\rm H}_n(1,1)+{\rm H}_n(2)$ & using the relationship above we make the sum equal to,

$\displaystyle J=2\zeta(3,1,1)+2\zeta(4,1)+\zeta(3,2)+\zeta(5)$

Calculating the multiple zeta values we get , $\displaystyle \sum_{n\ge 1} \frac{{\rm H}_{n}^{2}}{n^3} = \frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$

Next for $\text{**}$ , It can be shown similarly that $\displaystyle \sum_{n\ge 1} \frac{{\rm H}_{n}^{3}}{n^2} = 10\zeta(5)+\zeta(2)\zeta(3)$

Coming to the most complicated sum that is $\text{*}$ ,

We begin with writing $\displaystyle G=\sum_{n\ge 1}\frac{{\rm H}_n ({\rm H}_{n})^{(2)}}{n^2} = \sum_{n\ge 1}\frac{{\rm H}_{n-1} ({\rm H}_{n-1})^{(2)}}{n^2} + \sum_{n\ge 1} \frac{{\rm H}_{k-1}^{(2)}}{k^3}+\frac{{\rm H}_{k-1}}{k^4}+\zeta(5)$

Using MZV , $\displaystyle G = \sum_{n\ge 1}\frac{{\rm H}_{n-1} ({\rm H}_{n-1})^{(2)}}{n^2} + \zeta(3,2)+\zeta(4,1)+\zeta(5)$

Now using quasi-shuffle identity and multiple zeta ,

$\displaystyle \sum_{n\ge 1}\frac{{\rm H}_{n-1} ({\rm H}_{n-1})^{(2)}}{n^2} = \zeta(2,2,1)+\zeta(2,1,2)+\zeta(2,3)=2\zeta(2,3)+\zeta(3,2)$

Thus $\displaystyle \sum_{n\ge 1}\frac{{\rm H}_n ({\rm H}_{n})^{(2)}}{n^2} = \zeta(5)+\zeta(2)\zeta(3)$

Finally we derive , $\displaystyle \sum _{ k=1 }^{ \infty }{ \sum _{ j=1 }^{ \infty }{ \left[ \frac { { H }_{ j }\left( { H }_{ k+1 }-1 \right) }{ kj\left( k+1 \right) \left( j+k \right) } \right] } } = -\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)$ which makes the answer $\boxed{29}$

$\text{Note For MZV relations used above : }$ ,

$\zeta(4,1) = \zeta(5)-\zeta(3,2)-\zeta(2,3) \text{ (1)}$

$\begin{aligned} \zeta(4,1) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 y} \\ &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 (x-y)}, \text{ reindexing the second sum} \\ &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(-\frac{1}{x^4 y} - \frac{1}{x^3 y^2} - \frac{1}{x^2y^3} - \frac{1}{x y^4} + \frac{1}{(x-y)y^4}\right), \\ &\:\:\:\:\: \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(\frac{1}{(x-y)y^4} - \frac{1}{x y^4} \right) \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{y^4} \left(\frac{1}{x-y} - \frac{1}{x} \right) \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=y+1}^{\infty} \left(\frac{1}{x-y} - \frac{1}{x} \right), \\ & \:\:\:\:\: \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=1}^y \frac{1}{x}, \text{ as the sum telescopes} \\ &= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \zeta(4,1) + \zeta(5) \\ &= \zeta(5) - \zeta(3,2) - \zeta(2,3). \square \end{aligned}$

$\zeta(2,2,1)+\zeta(2,1,2)=\zeta(2,3)+\zeta(3,2) \text{ (2)}$

This is easy to see and rather a standard result so I'm skipping it.