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
Best Answer
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.