= Solution
Now $M_{n+1}=\max(M_n,X_{n+1})$. Again $X_{n+1}$ is <independent> of the whole previous history, so the recursion is a <Markov chain obtained from independent innovations>. To remain at $s$, the new observation must be at most $s$; to jump to $t>s$, it must equal $t$. Thus the <transition probabilities> are
$$
\boxed{P_{st}=\begin{cases}0,&t<s,\\\dfrac{s}{s+1},&t=s,\\\dfrac1{t(t+1)},&t>s.\end{cases}}
$$
The upper tail sums to $1/(s+1)$, complementing the holding <probability> $s/(s+1)$.
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