If $|a_{n+1}-a_n|\rightarrow0~(n\rightarrow\infty)$, the sequence itself does not necessarily have a limit (or not necessarily Cauchy). But must the sequence be bounded?
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Clarify the statement. "limit of two consecutive terms of a sequence goes to zero"? Individual terms don't have limits, – herb steinberg Feb 10 '19 at 04:09
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1No, it doesn't have to be bounded. Look at the harmonic numbers. – Matt Samuel Feb 10 '19 at 04:19
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Does this answer your question? Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero? – Rebecca J. Stones Jun 30 '22 at 06:53
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No, the sequence does not need to be bounded.
Consider, for example, $a_n=\sqrt{n}$. We have $$a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac1{2\sqrt{n}}\to 0$$ Of course, $\sqrt{n}\to \infty$. This sequence is not bounded.
jmerry
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Another, easy to remember example is $\log (n)$. $\log (n)$ tends to infinity, hence is not bounded. However, we have:
$$ |\log(n + 1) - \log (n)| = \left|\log \left( \frac{n+1}{n}\right)\right| $$ Which tends to $0$ as $n \to \infty$ .
rubikscube09
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