Conway and Guy defined $$H_k^0=\sum_{n=1}^k\dfrac1n$$ and $$H_k^r=\sum_{n=1}^kH_n^{r-1}$$ for $k,r\in\Bbb Z^+$.
I would prefer a definition of an $r$-hyperharmonic number to have some chance of converging when $r>0$. Consider that the harmonic series diverges because the integers are too slow-growing, but the harmonic numbers also begin at $1$ and grow much more slowly, so $$\sum_{n=1}^kH_n^0$$ doesn't grow much more quickly than the integers. My thinking is that we should consider the sum of the reciprocals of these sums, the sum of the reciprocals of the sums of those values, and so on. Formally, instead define
$$H_k^r=\sum_{n=1}^k\left(\sum_{m=1}^nH_m^{r-1}\right)^{-1}$$
Then implicitly, $H_k^{-1}=1$ for every positive integer $k$, as $$H_k^0=\sum_{n=1}^k\left(\sum_{m=1}^n1\right)^{-1}$$
My question is, does this series converge for all sufficiently large $r$?
Wolfram Alpha says the series diverges for $r=1$ by the comparison test. I figured I would compare the terms to the reciprocals of primes.
So we want to prove
$$\sum_{m=1}^nH_m^0\leq p_n$$
for sufficiently large $n$, where $p_n$ is the $n$th prime, but it is known that
$p_n>n\log n$, and
$$\sum_{m=1}^nH_m^0<n\log n$$
for $n\geq8$.
Any ideas for proving convergence/divergence for $r\geq2$?
Edit: I asked a separate question about closed forms for values of the function $$L_k=\lim_{r\to\infty}H_k^r$$ here, which is now resolved.
With the resolution of that question I believe $L_k$ can be defined in terms of $L_n$ for all $n<k$ via
$$U_k=\sum_{n=1}^k L_n=\left(\sqrt{\left(U_{k-1}+\sum_{n=1}^{k-1}\dfrac1{U_n}\right)^2+4}+U_{k-1}+\sum_{n=1}^{k-1}\dfrac1{U_n}\right)/2$$
as $L_k=U_k-U_{k-1}$, and I know now that
$$\lim_{k\to\infty}L_k=\infty$$
Considering that $U_k$ was so helpful in solving the other problem, perhaps we should generalize it to
$$U_k^r=\sum_{n=1}^kH_n^r,$$
so that
$$\lim_{r\to\infty}U_k^r=U_k$$
Then
$$U_k^r-U_{k-1}^r=\sum_{n=1}^k\dfrac1{U_n^{r-1}}$$
Unfortunately Sary's proof of the divergence of $(L_k)$ doesn't quite work here. I'm not sure if it can be fixed.
Edit: The formula for $U_k$ can be simplified slightly via
$$U_{k-1}+\sum_{n=1}^{k-1}\dfrac1{U_n}=2U_{k-1}-U_{k-2}$$