Prove that if a sequence is unbounded, then the sequence is not Cauchy

Question

So far:

If a sequence is unbounded, therefore it is a monotonic, divergent, sequence. Choosing $\epsilon = \frac12$, and $N\in\mathbb{N}$. Assume n, m $\geq N$. Then $\left| a_n - a_{N} \right| < \epsilon$ cannot be true, because...

I understand that if the sequence is unbounded, then the difference between $a_n$ and $a_N$ will eventually become greater than $\epsilon$, but I'm not sure how to write it out using the language of mathematics.

Answer 1

It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$\forall M \exists n: a_n>M$" or "$\forall M \exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+\epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>\epsilon$.

So this shows that for all $a_N$ and $\epsilon$, there is some $a_n$ that is more than $\epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+\epsilon$ for $n \leq N$.

Answer 2

By contradiction, let $(a_n)_{n\in\mathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|\leq1$ for $n\geq n_0$. But then $$ |a_n| \leq \max\{|a_1|,\dots,|a_{n_0-1}|,|a_{n_0}|+1\}, \quad \forall n\in\mathbb N, $$ so the sequence is bounded.