iterated harmonic numbers vs Riemann zeta

Question

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? $$\bar{H}_m(n)=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\binom{n}k.$$

Answer 1

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.

Answer 2

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.

Answer 3

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).