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I want to prove or disprove the following

Let $(a_n)$ be a bounded sequence such that $$\lim_{n\to\infty}(a_{n+1}-a_n)=0$$ Then $(a_n)$ is convergent.

My idea to prove that the proposition above is true is as follows: since $(a_n)$ is bounded, it has limit points. Then, by assuming that the sequence is not convergent, there are two different limit points $\alpha$ and $\beta$ and subsequences of $(a_n)$: $(a_{k_n})$ and $(a_{j_n})$ such that $a_{k_n}\to \alpha$ and $a_{j_n}\to\beta$. Now, I tried to build a subsequence $(a_{p_n})$ of $(a_n)$ with the property $p_{2n+1}=p_{2n}+1$ such that $(a_{p_{2n+1}})$ is a subsequence of $(a_{k_n})$ and $(a_{p_{2n}})$ is a subsequence of $(a_{j_n})$ thus there will exist a subsequence of the sequence $(a_{n+1}-a_{n})$ convergent to $\alpha-\beta$, which would be a contradiction. My problem was that I could not prove that you can always build the subsequece $(a_{p_n})$.

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No: consider the sequence $$1,0,\frac12,1,\frac23,\frac 13,0,\frac 14,\frac24,\frac 34,1,\frac 45,\frac35,\frac25,\frac15,0,\ldots$$

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  • $\begingroup$ Indeed, I was expenting the proposition to be wrong. $\endgroup$ – John WK Aug 23 '18 at 10:00
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So we know that it is false when we remove "bounded", just by letting $a_{n+1} - a_n = \frac1n$. This sequence "walks slowly to infinity"

We can produce a bounded example, just by setting up boundaries at say $-2$ and $2$ and reflecting the sequence as it progresses, so that it always stays within $[-2,2]$ but we still have $|a_{n+1}-a_n| = \frac1n$. Then this sequence doesn't converge.

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Moreover, we could choose some non convergent Cauchy sequence in a non-complete space, if we are not restricted to $\mathbb{R}$.

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