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Let $x_n$ be a sequence of real numbers in the interval $[a,b]$ for some $a,b \in \mathbb R$. If $x_n$ has the property: For all $\epsilon > 0$, there is $N \in \mathbb N$ such that for all $n\geq N: |x_{n+1} - x_n| < \epsilon$. Does this imply that the sequence converges?

I know that this is not true in general, for example, if $a_n = \sqrt n$ then $$a_{n + 1} - a_n \leq \frac{1}{2\sqrt n}$$ but maybe if the sequence is defined on a compact space this changes?

656475
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    No, and you can actually easily construct a counterexample from any counterexample on the reals by just periodically "flipping" it. Formally this is tedious to describe, but just have a zig-zag line function from the reals to that interval. If you don't follow I can elaborate and formalize this in a full answer. – kabel abel Jun 13 '24 at 23:47
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    To elaborate slightly on the previous comment, take partial sums of the harmonic series with the terms modified by a potential sign change so that the sum oscillates between $-1$ and $1$, never staying inside $[-1/2,1/2]$. – Ted Shifrin Jun 13 '24 at 23:57

3 Answers3

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Take for instance $a_n=\sin(2\ln(n))$

Then $a_{n+1}-a_n=2\underbrace{\sin\Big(\underbrace{\ln(n+1)-\ln(n)}_{O(\frac 1n)}\Big)}_{O(\frac 1n)}\underbrace{\cos\Big(\ln(n+1)+\ln(n)\Big)}_\text{bounded}\to 0$

Yet $a_n$ oscillates between $-1$ and $+1$ indefinitely.

zwim
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Take $x_n:=\sin(\sqrt n)$. This sequence is bounded between $-1$ and $1$ and meets your criterion, thanks to the inequality $$|\sin x - \sin y|\le |x-y|.$$ But $(x_n)$ doesn't converge because the sequence $(\sin (n))$ doesn't converge.

grand_chat
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Another example can be obtain from $x_n=1+\frac 1 2+ \cdots + \frac 1 n$. For any $p>0$ we have $x_{n+p}-x_n= \frac {1}{n+1}+ \cdots +\frac {1}{n+p}\leqslant \frac {p}{n+1}\to 0$, but $x_n$ is well known diverged sequence.

zkutch
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