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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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    $\begingroup$ I know that $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^2 = \frac{17}{4} \zeta(4).$$ Could you give us the expression for the $3$ case? By the way related paper. $\endgroup$ – user153012 Oct 27 '14 at 22:31
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    $\begingroup$ @user153012 $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3=\frac{31}{5040}\pi^6-\frac{5}{2}\zeta^2(3)=\frac{93}{16}\zeta(6)-\frac{5}{2}\zeta^2(3)$$ $\endgroup$ – user 1357113 Oct 27 '14 at 22:45
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    $\begingroup$ There is no closed form for the first Euler sum $s_h(4,4)$, but all weight-8 Euler sums can be expressed using zeta products and just one weight-8 Euler sum, whether it's $s_h(4,4)$ or $s_h(2,6)$ or some other one. Similarly for alternating Euler sums: there are four linearly independent alternating Euler sums necessary for a basis. (All this is conjectured.) $\endgroup$ – Kirill Oct 29 '14 at 13:21
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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

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  • $\begingroup$ What exactly is $S_{a,~b}$ ? $\endgroup$ – Lucian Apr 5 '17 at 15:14
  • $\begingroup$ The $S_{p,q}$ denotes the linear Euler sum or double zeta value defined by $\endgroup$ – xuce1234 Apr 5 '17 at 23:00
  • $\begingroup$ The $S_{p,q}$ denotes the linear Euler sum or double zeta value defined by$$S_{p,q}:=\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( p \right)}}} {{{n^q}}}}, $$ where $p,q$ are positive integers with $q \geq 2$, and $w:=p+q$ denotes the weight of linear sums $S_{p,q}$. $\endgroup$ – xuce1234 Apr 5 '17 at 23:02
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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.

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  • $\begingroup$ Can anyone help me clean up the LaTeX? $\endgroup$ – Aditya Kumar Jul 6 '16 at 16:48
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    $\begingroup$ Interesting approach using MZV (+1) $\endgroup$ – user 1357113 Jul 6 '16 at 19:14

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