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Prove or disprove that $|a_{n+1}-a_n|<1/n$, then ${a_n}$ is a Cauchy sequence.

I encountered a similar question the other day, that question is about $|a_{n+1}-a_n|<\frac{1}{n^2}$, then ${a_n}$ is a Cauchy sequence. I noticed that that one can be related to series $\displaystyle\sum_{n=1}^\infty{\frac{1}{n^2}}$, and this series is convergent. However, $\displaystyle\sum_{n=1}^\infty{\frac{1}{n}}$ is divergent, so I guess the answer is no. But if it is false, I need to provide a explicit counterexample.

Can anyone help me? Thanks in advance!

Sourav Ghosh
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Lumos
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    No, this series is divergent. Because you sum on $k$, not on $n$. $$\sum_{k=1}^\infty \frac{1}{n^2}\to\infty$$ – MathFail May 12 '23 at 03:27

1 Answers1

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Hint:

A sequence $(a_n)$ such that $d(a_{n+1}, a_n) \to 0$ as $n\to\infty$ is called quasi-Cauchy sequence.

Exercise:

  1. Every Cauchy sequence is quasi-Cauchy.

  2. A quasi-Cauchy sequence need not be Cauchy $($Hint: $(a_n) =\sqrt{n}$ or $a_n=\log n)$

Your observation is correct✅

A quasi-Cauchy sequence $(a_n) $ where $d(a_{n+1}, a_n) $ can be dominated by the general term of a convergent series then the quasi-Cauchy would be a Cauchy sequence.

See here for the proof.

Sourav Ghosh
  • 12,997