Let there be a sequence such that the distance between two consecutive terms converges to 0. Must this sequence converge?

Question

I'm trying to solve the following analysis problem and I've developed a proof, I'm just not entirely sure if it's valid or not.

Let $\{a_n\}_n$ be a sequence such that $|a_n -a_{n+1}|\to 0$. Must $\{a_n\}_n$ converge? If so prove it, and if not, find a counterexample.

Proof:

Let $\epsilon > 0$. Since $|a_n -a_{n+1}| \to 0$, there exists $N\in \Bbb R$ such that$|a_n-a_{n+1}| \lt \frac {\epsilon}{m+n}$for all $n > N$. Then, for all $m,n > N$, we have $$|a_m - a_n|= |a_m - a_{m-1} + a_{m-1} - a_{m-2} + ... + a_{n+1} - a_{n}| $$ $$\le |a_m - a_{m-1}| + |a_{m-1} - a_{m-2}| + ... + |a_{n+1} - a_n|$$ $$ = \sum_{n}^{m - 1}{|a_{i+1}-a_{i}|} = \sum_{n}^{m - 1}{|a_{i}-a_{i+1}|}$$ $$ \lt \sum_n^{m-1} {\frac {\epsilon}{m+n}} = \frac{\epsilon\;(m+n)}{m+n}=\epsilon.$$ Therefore, the sequence is Cauchy and converges. $\Box$

Any help with checking/improving my proof would be greatly appreciated. Thank you!

Your choice of $N$ depends on $m$ and $n$; there may be no integers $n,m >N$! — Kavi Rama Murthy, Mar 07 '19 at 07:38
@KaviRamaMurthy But isn't it always guaranteed that there is an integer greater than any given real number by the Archimedean Property? — BalsamicVinegar, Mar 07 '19 at 07:41
Ah, I think I see the issue. So when I'm doing a convergence type proof like this, can I only have my N depend on epsilon? — BalsamicVinegar, Mar 07 '19 at 07:46

score 4 · Accepted Answer · answered Mar 07 '19 at 07:37

4

$a_n=1+\frac 1 2+\cdots+\frac 1 n$ is a counterexample.

answered Mar 07 '19 at 07:37

Kavi Rama Murthy

359,332

Let there be a sequence such that the distance between two consecutive terms converges to 0. Must this sequence converge?

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