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



iterated harmonic numbers vs Riemann zeta
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](http://www.dms.umontreal.ca/~andrew/PDF/Riemann.pdf) 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. 


