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If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.

However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| < \frac{\epsilon}{m-n}$

Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|<|a_m -a_n| <\epsilon$

This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.

But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks

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  • $\begingroup$ Already you have a counter example. $\endgroup$ – Mhenni Benghorbal Apr 25 '14 at 6:17
  • $\begingroup$ This is really not Cauchy, as you said; you need to have a value $k$ , so that $|a_n-a_m|< \epsilon$ for all $n,n >k$ , an dnot just for $n=m+1$. $\endgroup$ – user99680 Apr 25 '14 at 6:17
  • $\begingroup$ what if i have an additional condition of boundedness, is a_n convergent then? $\endgroup$ – user10024395 Apr 25 '14 at 6:18
  • $\begingroup$ See this result. $\endgroup$ – Mhenni Benghorbal Apr 25 '14 at 6:25
  • $\begingroup$ this is actually referring to a different thing, I am saying if a_n itself is bounded not the sum of |an-am| is bounded. $\endgroup$ – user10024395 Apr 25 '14 at 6:30
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Even if the sequence is bounded, the condition does not imply that the sequence is Cauchy.

Consider the following sequence: $$ 0,1,\frac12,0,\frac14,\frac12,\frac34,1,\frac78,\frac68,\frac58,\frac48,\frac38,\frac28,\frac18,0,\frac1{16},\ldots $$ The sequence goes back and forth between $0$ and $1$ in smaller and smaller steps. So $\lim(a_{n+1}-a_n)=0$, while the sequence oscillates between $0$ and $1$ and so it is not Cauchy.

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Take any series $\sum b_n$ of positive number, which diverges, but such that $b_n\to0$. Then we can form a new series $\sum \varepsilon_nb_n$, with $\varepsilon=\pm1$ (i.e., it is the same series with possible change of signs) in a such way the for the partial sums $a_n=\sum\limits_{k=1}^n \varepsilon_kb_k$ we have $$-\infty < \liminf\limits_{n\to\infty} a_n<\limsup\limits_{n\to\infty} a_n <+\infty.$$ (In fact, the limit superior and limit inferior can be chosen arbitrarily. This is related to unconditional convergence)

Now for the sequence $(a_n)$ you have $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ and this sequence is bounded. Yet it is a sequence of real numbers which is not convergent and thus not Cauchy.

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  • $\begingroup$ why is the an bounded here? i can't really see why $\endgroup$ – user10024395 Apr 25 '14 at 6:49
  • $\begingroup$ @user136266 Any sequence with $\limsup a_n<+\infty$ is bounded from above. If we denote $\overline L=\limsup a_n$, then $a_n<\overline L+\varepsilon$ for each $n>n_0$. The finitely many values $a_1,a_2,\dots,a_{n_0}$ cannot influence boundedness. We can show similarly that if $\liminf a_n>-\infty$ then the sequence is bounded from below. $\endgroup$ – Martin Sleziak Apr 25 '14 at 7:02
  • $\begingroup$ why i can be sure that it has a lim sup? $\endgroup$ – user10024395 Apr 25 '14 at 7:06
  • $\begingroup$ @user136266 I have chosen in that way that it has a finite $\limsup$. (I have mentioned a result in the first paragraph saying that the signs can be chosen in such a way that both limsup and liminf are finite. I cannot find a reference for this result at the moment, but the basic idea of the proof is not that difficult.) $\endgroup$ – Martin Sleziak Apr 25 '14 at 7:12
  • $\begingroup$ @user136266 I have posted more precise formulation of this result and sketch of the proof in a different post. However, I should say that my answer is unnecessary complicated - the answer posted by Martin Argerami gives a counterexample which is much easier to understand. $\endgroup$ – Martin Sleziak Apr 25 '14 at 7:58

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