= Solution
Let $Q=H_N$ and define
$$
\pi_N=\frac1Q,\qquad
\pi_k=\frac1{(k+1)Q}\quad(1\leq k<N).
$$
The probabilities sum to $Q^{-1}(1+\sum_{k=1}^{N-1}(k+1)^{-1})=1$. For $k<N$, the incoming <probability> is
$$
(\pi P)_k=\sum_{j=k+1}^{N}\frac{\pi_j}{j}
=\frac1Q\left(\sum_{j=k+1}^{N-1}\frac1{j(j+1)}+\frac1N\right)
=\frac1{(k+1)Q}=\pi_k,
$$
using a telescoping sum. Similarly,
$$
(\pi P)_N=\sum_{j=1}^N\frac{\pi_j}{j}
=\frac1Q\left(\sum_{j=1}^{N-1}\frac1{j(j+1)}+\frac1N\right)
=\frac1Q=\pi_N.
$$
Thus $\pi$ is a <stationary distribution>. Every state can reach $N$ in one step, and $N$ can reach every lower positive state in one step, so the chain is <irreducible Markov chain>[irreducible]. The self-loop at $N$ has <probability> $1/N>0$, hence the chain is <aperiodic Markov chain>[aperiodic]. The convergence theorem for a finite irreducible aperiodic <Markov chain> now gives
$$
\boxed{\lim_{n\to\infty}P(X_n=k)=\pi_k.}
$$
For $N=1$, the conclusion follows directly from the single absorbing/refilled state.
This is the limiting law of the <uniform depletion and refill Markov chain>.
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