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Define the $m$-th iterated harmonic sums in the manner: $\bar{H}_0(n):=1$ and for $m\geq1$ by $$\bar{H}_m(n):=\sum_{k=1}^n\frac{\bar{H}_{m-1}(k)}k.$$ For example, $\bar{H}_1(n)=\sum_{k=1}^n\frac1k$ are the familiar harmonic numbers. Euler proved that $$\frac12\sum_{n\geq1}\frac{\bar{H}_1(n)}{n^2}=\zeta(3).$$

Hoping for a natural generalization, I ask:

Question 1. Is this true? If so, any proof? $$\frac1{m+1}\sum_{n\geq1}\frac{\bar{H}_m(n)}{n^2}=\zeta(m+2).$$

Of course, this works for $m=0$ as well: $\frac1{0+1}\sum_{n\geq1}\frac{\bar{H}_0(n)}{n^2}=\zeta(2)$.

Question 2. This might be of auxiliary interest. Any proof? $$\bar{H}_m(n)=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\binom{n}k.$$

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    $\begingroup$ how far did you check it numerically? $\endgroup$ – ARG Jun 18 '17 at 18:23
  • $\begingroup$ Quite extensively. $\endgroup$ – T. Amdeberhan Jun 18 '17 at 18:28
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    $\begingroup$ If true, it would be a really beautiful formula. $\endgroup$ – Sylvain JULIEN Jun 18 '17 at 18:29
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    $\begingroup$ This infinite sum can be expressed in terms of multiple zeta values. In the case $m=2$ we have $\sum_{n \geq 1} H_2(n)/n^2 = \zeta(2,1,1)+\zeta(3,1)+\zeta(4)+\zeta(2,2) = \zeta(4)+(1/4)\zeta(4)+\zeta(4)+(3/4)\zeta(4)=3\zeta(4)$ as you predicted (these relations date back essentially to Euler). $\endgroup$ – François Brunault Jun 18 '17 at 18:57
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    $\begingroup$ Regarding Question 2, I am quite certain the identity is a special case of an identity of Hoffman and Kawashima. See arxiv.org/abs/math/0702824 and arxiv.org/abs/math/0401319 $\endgroup$ – GH from MO Jun 18 '17 at 19:04
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Answer to Question 1.

Define the multi-zeta value $\zeta(p_1,\ldots, p_g)$ as follows: $$ \zeta(p_1,\ldots,p_g) = \sum_{a_1>a_2 >\ldots > a_g\ge 1}\frac{1}{a_1^{p_1}\cdots a_g^{p_g}}, $$ where $p_1\ge 2$ and the other $p_j$ are integers $\ge 1$. Granville and Zagier independently showed that $$ \sum_{p_1+\ldots+p_g= N} \zeta(p_1,\ldots,p_g) = \zeta(N), $$ where in the sum $g$ is fixed, and the sum is over all tuples with $p_1\ge 2$ and other $p_j \ge 1$. This generalizes Euler's relation $$ \zeta(2,1)=\zeta(3). $$ Your result follows from this, upon distinguishing in your harmonic sum when some terms can equal others. For example consider your $m=2$ case which is $$ \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1\le n} \frac{1}{m_1} \sum_{m_2\le m_1}\frac{1}{m_2} = \sum_{n=1}^{\infty} \frac{1}{n^3} \sum_{m_2 \le n} \frac{1}{m_2} + \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1 <n} \frac{1}{m_1} \sum_{m_2 \le m_1} \frac{1}{m_2} $$ and the first term is $$ \zeta(4)+\zeta(3,1), $$ while the second term is $$ \zeta(2,2)+\zeta(2,1,1). $$ These add up to $3\zeta(4)$ by Granville-Zagier.

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    $\begingroup$ Nice proof. Actually, the OP's left hand side is the multiple zeta star value $\zeta^\ast(2,1_m)$, so chances are that his identity is already in the literature. $\endgroup$ – GH from MO Jun 18 '17 at 18:41
  • $\begingroup$ Lucia: great reference. Compiling all the places wherever equality occurs (such as your examples, $\zeta(2,2), \zeta(2,1,1),$ etc requires additional justification, especially in higher dimensions. What do you do there? $\endgroup$ – Lewi_Sol Jun 18 '17 at 18:52
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    $\begingroup$ @Lewi_Sol: In expanding out the harmonic sums, all that matters is how many of them are distinct. Since there are $m+1$ terms (including $n$) this gives $m+1$ possible combinations. $\endgroup$ – Lucia Jun 18 '17 at 18:54
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    $\begingroup$ @T. Amdeberhan: I don't mind your edit, but when I answered the question, there was only Question 1! $\endgroup$ – Lucia Jun 18 '17 at 19:05
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    $\begingroup$ See also my response. $\endgroup$ – GH from MO Jun 18 '17 at 19:22
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The identity in Question 1 is a special case of Theorem 2 in Ohno: A Generalization of the Duality and Sum Formulas on the Multiple Zeta Values (see here). Indeed, putting $k=1$ and $n=m+1$ in this theorem, and noting that $\xi_1(s)$ equals $s\zeta(s+1)$, the result follows.

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    $\begingroup$ That's a nice generalization. Minor comment: I guess in the last line you want $\xi_1(s)$ in Ohno's notation. $\endgroup$ – Lucia Jun 18 '17 at 20:33
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Unless I messed up the notation, it seems that Note 5.3 of this paper

Luis A. Medina, Victor H. Moll, and Eric S. Rowland, Iterated primitives of logarithmic powers, International Journal of Number Theory 7 (2011) pp 623–634, doi:10.1142/S179304211100423X, arXiv:0911.1325 (pdf)

answers Question 2.

It seems quite likely that the OP is well aware of the above paper, given this recent paper:

Tewodros Amdeberhan, Christoph Koutschan, Victor H. Moll, Eric S. Rowland, The iterated integrals of $\ln(1 + x^n)$, International Journal of Number Theory 8 (2012) pp 71–94 doi:10.1142/S1793042112500042, arXiv:1012.3429 (.ps file).

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