Can we construct a divergent sequence where the difference between consecutive terms is less than epsilon for $n \geq N$?

Question

Can we find examples of sequences $a_n$ where $a_n$ is divergent and for any $\epsilon > 0$, there exists an $N \in \mathbf{N}$ such that $|a_{n+1} - a_n| < \epsilon$ for all $n \geq N$?

My initial guess is that we can construct some sequence that oscillates between two values whose difference can always be made to be less than $\epsilon$. Something like $a_n = \frac{n}{n + 2}$ for odd $n$, $a_n = \frac{n + 1}{n+2}$ for even $n$. Does this seem right?

If so, I think this is a pretty cool way to satisfy both conditions, but it would be even cooler if we could do it with a monotonic sequence. Is there some monotonic $a_n$ that satisfies both of these constraints? Thanks for your time!

Answer 1

Yes, we can. Try $a_n = \ln n$, for instance.

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Equivalently, what you ask for is "is there a (preferably monotonic) divergent sequence $(a_n)_n$ such as $a_{n+1}-a_n \xrightarrow[n\to\infty]{}0$?"

Again equivalently, what you ask for is "is there a (preferably positive) divergent series $\sum_n a_n$ such as $a_n \xrightarrow[n\to\infty]{}0$?"

The answer to both is yes.

Answer 2

How about $a_n=\sum\limits_{k=1}^n\frac1k$?

Answer 3

$a_n=√n$;

$a_{n+1}-a_n=\sqrt{n+1}-√n=$

$\dfrac{1}{\sqrt{n+1}+√n}< \dfrac{1}{2√n}$.