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 $x_n=x_k, \forall n \ge k$.
 A: 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 $n\ge k$.
A: 
We will first show that such a sequence $(x_n)$ of natural numbers satisfy:  $\forall n\in\mathbb{N}$,
  if $x_{n+1} < x_{n}$, then $x_{n+1} > n$.

$(n+1)x_{n+1} - nx_n > 0$ can also be written as $n(x_{n+1}-x_n) + x_{n+1} > 0$. 
If $x_{n+1} < x_{n}$, then the above condition gives us $n(-1) + x_{n+1} \geq n(x_{n+1}-x_n) + x_{n+1} > 0$ which implies $x_{n+1} > n$. Therefore, either $x_{n+1}\geq x_n$ or $x_{n+1} > n$.

Now we will show that if the sequence $(x_n)$ is bounded, then there exists $k\in\mathbb{N}$, such that $x_n = x_k$ $\forall n\geq k$.

Suppose $B\in\mathbb{N}$ is the smallest upper bound of the sequence $(x_n)$ of natural numbers, then there exists $l\in \mathbb{N}$ such that $x_l = B$. 
If $l\geq B$, then ($x_{l+1}\geq B$ or $x_{l+1} > l$) implies that $x_{l+1} = B$. Inductively, we get that for $l$, $x_n = x_l = B$ $\forall n\geq l$.
If $l < B$, then $\forall \ t\geq B$, $x_{t+1}\geq x_t$. Since it is monotonic and bounded, and only consists of natural numbers, there exists $k\geq B$, such that $x_n = x_k$ $\forall n\geq k$.
