Problem involving sequence and inequality Let $(x_n)_{n \ge 1}$ be a sequence of natural numbers such that $(n+1)x_{n+1}-nx_n \gt 0$. Show that, if the sequence is bounded, than there exists $k \in \Bbb N, k \ge 1$ such that $x_n=x_k, \forall n \ge k$.

Since all $(x_n)$ are natural numbers, the given assumption $(n+1)x_{n+1}-nx_n>0$ can be rewritten as
$$x_{n+1}>\frac{n}{n+1}x_n=\left(1-\frac{1}{n+1}\right)x_n=x_n-\frac{x_n}{n+1}.$$
If the sequence is bounded, then there exists $M\in\mathbb{N}$ such that all $x_n\le M$. Then for any $n\ge M$:
$$\frac{x_n}{n+1}\le\frac{M}{M+1}<1 \quad \implies \quad x_{n+1}>x_n-\frac{x_n}{n+1}>x_n-1 \quad \implies \quad x_{n+1}\ge x_n.$$
the last inequality being true because the terms in the sequence are integers.
If $x_{n+1}>x_n$ holds for infinitely many $n\ge M$, then the sequence would be unbounded (all terms are integers, so every time the sequence grows, it grows at least by one), contradicting the assumption. Therefore $x_{n+1}>x_n$ holds only for finitely many $n\ge M$. Let $k$ be the index of the larger element (i.e. $k=n+1$) in the last pair for which $x_{n+1}>x_n$. Then $x_{n+1}=x_n$ for all $n\ge k$.
