Let $(x_n)_{n\in\ \mathbb N}$ and $(y_n)_{n\in\ \mathbb N}$ be two sequences in $\mathbb R$. Suppose $|x_n – x_m| \le y_n$ for all $n,m \in \mathbb N$ with the condition that $m \ge n$. Show that if $\lim_{n \rightarrow \infty} y_n = 0$, then $(x_n)_{n\in\ \mathbb N}$ is a convergent.
I'm guessing that I should most definitely prove the sequence $(x_n)_{n \in \mathbb N}$ is a cauchy sequence and from there use the big Cauchy theorem.
Since we can assume the $\lim_{n \rightarrow \infty} y_n = 0$, I know for any $\epsilon >0$ there exists $n\in \mathbb N$, then $\forall m \in \mathbb N$ such that $m \ge n$ we have $|x_n – x_m| \le |y_n| \le \epsilon$, but I'm stuck here since, I need to show that $|x_n – x_m| \le \epsilon$ $\forall n,m \in \mathbb N$ and choose a new $N \in \mathbb N$ such that these are greater than N.
I'm wondering if I could get any feedback, thank you.
Best Answer
Given $\epsilon> 0$, there exists $N$ such that $n\geq N$ implies that $|y_n|<\varepsilon$. In particular if $m\geq n\geq N$, then $$ |x_n-x_m|\leq y_n<\varepsilon, $$ This means that $(x_n)$ is a cauchy sequence and thus converges.