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. 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: $\frac1{0+1}\sum_{n\geq1}\frac{\bar{H}_0(n)}{n^2}=\zeta(2)$.

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

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

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. 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: $\frac1{0+1}\sum_{n\geq1}\frac{\bar{H}_0(n)}{n^2}=\zeta(2)$.