# Cauchy sequences. Show that $(x_n)$ is Cauchy.

Let $(x_n)$ and $(y_n)$ be sequences such that $\lim y_n = 0$. Suppose that for all $k \in \Bbb N$ and all $m ≥ k$ we have $|x_m − x_k| ≤ y_k$. Show that $(x_n)$ is Cauchy.

I need a little guidance on how to approach the problem. As I see this is the same definition of Cauchy sequences. But I do not see how to connect everything in a logic sequence in order to have a rigorous proof.

My attempt of reasoning

I started first defining the $\lim$ of $y_n$.

For every $\varepsilon>0$ exists $N$ s.t. $n>N$ $|y_n|<\varepsilon$ for all $n>N$ Then I see that all terms of $y_n$ get smaller and smaller as $n$ gets larger. So distance between $x_m$ and $x_k$ gets smaller as the terms get bigger. But one thing that puts me off is that $m ≥ k$ and $| x_m − x_k| ≤ y_k$ why are they $\leq$?

Thanks for help in advance'

## 1 Answer

I think the way you started out is great. We need to show the following:

Given $\epsilon > 0$, there exists $N$ such that $n, m \geq N$ implies $|x_{n} - x_{m} | < \epsilon$.

But, as you correctly noted, given $\epsilon > 0$, $\exists N$ such that $n \geq N$ implies $|y_{n}| < \epsilon$.

But we have the inequality that for all $m \geq n$, $|x_{m} - x_{n} | \leq |y_{n}|$ (actually, this inequality also holds for all $n$, and in particular, for all $n \geq N$), which means $|x_{m} - x_{n} | < \epsilon$ for all $m \geq n$ (and all $n \geq N$). In other words, for all $n, m \geq N$, $|x_{m} - x_{n}| \leq |y_{n}| < \epsilon$.

So, given $\epsilon > 0$, we found our $N$ (the $N$ that made $|y_{n}| < \epsilon$ if $n \geq N$) such that $n, m \geq N$ implies $|x_{n} - x_{m}| < \epsilon$, which is exactly the definition of the sequence being Cauchy.

• @MarussT You're welcome! By the way, if you receive an answer to your question that you think is the "best" one, or that answers all of your questions, you should close your question by choosing the answer as best (clicking the check mark next to the answer you want to choose as best). If you want to possibly wait for other others, that's OK too! – layman Sep 29 '14 at 22:14