Counterexample for $\lim_{n\to\infty} (a_{n-1}-a_n) = 0 \Rightarrow a_n$ converges

Question

I think the equation above is false, but I cannot find a counterexample.

Also, I think that the implication is true when $a_n^2 \to 1$ is also required.

Suppose we have $a_n\in\mathbb{C}\ \forall n\in\mathbb{N}$.

Can you help me?

score 5 · Accepted Answer · answered Feb 08 '19 at 18:55

5

Hint: What happens if $a_n=1+\frac12+\frac13+\cdots+\frac1n$?

answered Feb 08 '19 at 18:55

José Carlos Santos

427,504

Thank you. I've forgotten that we can also consider partial sums of a series. – Aericura Feb 08 '19 at 19:03

score 2 · Answer 2 · answered Feb 08 '19 at 18:58

2

If $a_n=\log n,$ $\lim_{n\rightarrow \infty}(a_n-a_{n-1})=\lim_{n\rightarrow \infty}\log (1+\frac1{n-1})=0$. However, $a_n\rightarrow \infty (n\rightarrow\infty).$

answered Feb 08 '19 at 18:58

Kenta S

16,151
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Counterexample for $\lim_{n\to\infty} (a_{n-1}-a_n) = 0 \Rightarrow a_n$ converges

2 Answers2