Let $(x_n)$ be a sequence s.t l. $|x_m - x_n|\le\frac1{1+|m-n|}$ for all $m$ and $n$. Find (with proof) $\lim x_n$.

Question

Suppose $(x_n)$ is a sequence that satisfies

$$ |x_m - x_n| \leq \dfrac{1}{1+|m-n|} $$

for all $m$ and $n$. Find (with proof) $\lim x_n$

ATTEMPT:

If we put $m = n+1$, then

$$ |x_{n+1} - x_n | \leq \dfrac{1}{1+n} $$

From this condition we see that $\lim x_n = L$ exists. On the other Hand, if we choose $n=m-1$, we obtain

$$ |x_{m} - x_{m-1} | \leq \dfrac{1}{2} $$

And, as $m$ can by any number, then we may not conclude that the sequence converges as it may oscillate. Can I get some hints on how to solve this? Is my reasoning reasonable? or am I overthinking and it is a simple exercise.

Answer 1

HINT: Fix $m\in\Bbb N$. Then for each $n>m$ we have

$$|x_n-x_m|\le\frac1{1+(n-m)}=\frac1{n-(m-1)}\,,$$

which tends to $0$ as $n$ increases without bound. Thus, $\lim\limits_{n\to\infty}x_n=x_m$. This is true for each $m\in\Bbb N$, so ... ?