Closed form of $\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$

Solution 1:

The result is \begin{align*} S_{1^4,4}=\sum\limits_{n = 1}^\infty {\frac{{H_n^4}}{{{n^4}}}} = \frac{{13559}}{{144}}\zeta \left( 8 \right) - 92\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 26{S_{2,6}}, \end{align*} Detailed process see the paper ``Multiple zeta values and Euler sum" enter link description here

Solution 2:

First I'll simplify $\displaystyle S={ \left( \sum _{ m=1 }^{ n }{ \frac { 1 }{ m } } \right) }^{ 4 }$

Now, we can re-write it as: $$S=\left( \sum _{ { m }_{ 1 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 1 } } } \right) \left( \sum _{ { m }_{ 2 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 2 } } } \right) \left( \sum _{ { m }_{ 3 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 3 } } } \right) \left( \sum _{ { m }_{ 4 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 4 } } } \right) $$

This satisfies quasi-shuffle identity. So I can re-write it as:

$\displaystyle S=\left[ \underbrace{\left( \sum _{ n\ge { m }_{ 1 }>{ m }_{ 2 }>{ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 1 }>{ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \cdots \right)}_{24 \ terms} +\underbrace{\left( \sum _{ n\ge { m }_{ 1 }>{ m }_{ 2 }>{ m }_{ 3 }={ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 1 }>{ m }_{ 3 }={ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }>{ m }_{ 3 }>{ m }_{ 2 }={ m }_{ 4 }>1 }^{ }{ + } ... \right)}_{12 \ terms} +\underbrace{\left( \sum _{ n\ge { m }_{ 1 }>{ m }_{ 2 }={ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 1 }={ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }>{ m }_{ 4 }={ m }_{ 2 }>{ m }_{ 4 }>1 }^{ }{ + } \cdots \right)}_{12 \ terms} +\underbrace{\left( \sum _{ n\ge { m }_{ 1 }={ m }_{ 2 }>{ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }={ m }_{ 3 }>{ m }_{ 1 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }={ m }_{ 4 }>{ m }_{ 2 }>{ m }_{ 3 }>1 }^{ }{ + } \cdots \right)}_{12 \ terms} +\underbrace{\left( \sum _{ n\ge { m }_{ 1 }={ m }_{ 2 }={ m }_{ 3 }>{ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }={ m }_{ 4 }={ m }_{ 2 }>{ m }_{ 4 }>1 }^{ }{ + } \cdots \right)}_{4 \ terms} +\underbrace{\left( \sum _{ n\ge { m }_{ 1 }>{ m }_{ 2 }={ m }_{ 3 }={ m }_{ 4 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 4 }={ m }_{ 1 }={ m }_{ 3 }>1 }^{ }{ + } \cdots \right)}_{4 \ terms} +\sum _{ n\ge { m }_{ 1 }={ m }_{ 2 }={ m }_{ 3 }={ m }_{ 4 }>1 }^{ }{ } \right] \frac { 1 }{ { m }_{ 1 }{ m }_{ 2 }{ m }_{ 3 }{ m }_{ 4 } } $

Now, on exploiting symmetry and using multi-harmonic sum I'll re-write it as: $$S=24{ H }_{ n }\left( 1,1,1,1 \right) +12{ H }_{ n }\left( 1,1,2 \right) +12{ H }_{ n }\left( 1,2,1 \right) +12{ H }_{ n }\left( 2,1,1 \right) +4{ H }_{ n }\left( 1,3 \right) +4{ H }_{ n }\left( 3,1 \right) +{ H }_{ n }\left( 4 \right) $$

Now, coming back to the problem. I'll insert the value of $S$ here.

$\displaystyle A=\sum _{ n=1 }^{ \infty }{ \frac { 24{ H }_{ n }\left( 1,1,1,1 \right) +12{ H }_{ n }\left( 1,1,2 \right) +12{ H }_{ n }\left( 1,2,1 \right) +12{ H }_{ n }\left( 2,1,1 \right) +4{ H }_{ n }\left( 1,3 \right) +4{ H }_{ n }\left( 3,1 \right) +{ H }_{ n }\left( 4 \right) }{ { n }^{ 4 } } } $

Now, I'll use the relation of multi-harmonic sum and multi-zeta variable: $$\sum_{n=1}^\infty \frac{H_n(s_1,\ldots,s_k)}{n^s}=\zeta(s,s_1,\ldots,s_k)+\zeta(s+s_1,s_2,\ldots,s_k).$$

Therefore, the final answer is

$\boxed{A=24\zeta \left( 4,1,1,1,1 \right) +24\zeta \left( 5,1,1,1 \right) +12\zeta \left( 4,1,1,2 \right) +12\zeta \left( 5,1,2 \right) +12\zeta \left( 4,1,2,1 \right) +12\zeta \left( 5,2,1 \right) +12\zeta \left( 4,2,1,1 \right) +12\zeta \left( 6,1,1 \right) +4\zeta \left( 4,1,3 \right) +4\zeta \left( 5,3 \right) +4\zeta \left( 4,3,1 \right) +4\zeta \left( 7,1 \right) +\zeta \left( 4,4 \right) +\zeta \left( 8 \right) }$

Note that further simplification of certain MZVs is very tough to do by hand.