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

Question

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

I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a
closed form for this variant. Is it possible to tackle the question in an elementary way
and find the answer, without using integrals at all?

Then, if this exists, I'd also propose the alternating variant

$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\frac{H_n}{n}\right)^4$$

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Update (by editor): The last sum by OP is: $$\small \sum _{n=1}^{\infty } (-1)^{n-1} \left(\frac{H_n}{n}\right){}^4=\frac{633}{128} \zeta(6,2)-\frac{1}{6} \pi ^2\zeta(\bar5,1)+6 \zeta(\bar7,1)-4 \zeta(\bar5,1,\bar1,1)+6 \log ^2(2) \zeta(\bar5,1)+12 \log (2) \zeta(\bar5,1,1)-16 \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)-\frac{1}{9} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^2 \zeta (3)^2}{8}-\frac{63 \zeta (3) \zeta (5)}{64}+\frac{2}{15} \zeta (3) \log ^5(2)-\frac{2}{9} \pi ^2 \zeta (3) \log ^3(2)-\frac{31}{12} \zeta (5) \log ^3(2)+3 \zeta (3)^2 \log ^2(2)-\frac{16}{45} \pi ^4 \zeta (3) \log (2)-\frac{65}{48} \pi ^2 \zeta (5) \log (2)+\frac{155}{8} \zeta (7) \log (2)+\frac{149141 \pi ^8}{43545600}-\frac{1}{216} \pi ^4 \log ^4(2)+\frac{25 \pi ^6 \log ^2(2)}{1512}$$

Answer 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

Answer 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.