Boundedness and Cauchy Sequence: Is a bounded sequence such that $\lim(a_{n+1}-a_n)=0$ necessarily Cauchy?

Question

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

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 $n=m+1$. — user99680, Apr 25 '14 at 06:17
what if i have an additional condition of boundedness, is a_n convergent then? — user10024395, Apr 25 '14 at 06:18
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. — user10024395, Apr 25 '14 at 06:30
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. — Martin Sleziak, Apr 25 '14 at 06:37
In this case I have edited your title, please have a look whether you are satisfied with the new title, end if not, please edit it further to improve it. — Martin Sleziak, Apr 25 '14 at 06:38

score 4 · Accepted Answer · answered Apr 25 '14 at 06:31

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.

Martin Sleziak · Answer 2 · 2014-04-25T06:39:35.930

0

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.

edited Apr 25 '14 at 06:39

answered Apr 25 '14 at 06:33

Martin Sleziak

53,687

why is the an bounded here? i can't really see why – user10024395 Apr 25 '14 at 06:49
@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. – Martin Sleziak Apr 25 '14 at 07:02
why i can be sure that it has a lim sup? – user10024395 Apr 25 '14 at 07:06
@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.) – Martin Sleziak Apr 25 '14 at 07:12
@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. – Martin Sleziak Apr 25 '14 at 07:58

Boundedness and Cauchy Sequence: Is a bounded sequence such that $\lim(a_{n+1}-a_n)=0$ necessarily Cauchy?

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