Is it true that for each properly divergent sequence $(a_n)$, $\exists k \in \Bbb N$ such that $\sum_1^{\infty} 1/{a_n}^{k}$ is convergent?

Question

Let $(a_n)$ be a given properly divergent sequence (meaning $(a_n) \to \infty$). Does there always exist some $k \in \Bbb N$ such that $\sum_{n=1}^{\infty} {1/{a_n}}^{k}$ is a convergent series?

It's a conjecture that I couldn't disprove(or prove). I am sorry if I have just asked something that is too obvious.

Edit : Please assume $a_n \ne 0 \ \forall \ n \in \Bbb N$

score 2 · Accepted Answer · answered Dec 02 '16 at 20:25

2

No, this is not always the case. You can take $a_n=\ln(n),n \geq 2$. See for instance here.

answered Dec 02 '16 at 20:25

Watson

24,404

Is it true that for each properly divergent sequence $(a_n)$, $\exists k \in \Bbb N$ such that $\sum_1^{\infty} 1/{a_n}^{k}$ is convergent?

1 Answers1