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

Answer 1

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.

Answer 2

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.