$\lim_{n\to\infty}na_n=0\Rightarrow\sum_{n=1}^\infty a_n<\infty$?

Question

Let $\{a_n\}_{n\in\mathbb{N}}\subset\mathbb{R}$ be a nonnegative decreasing sequence satisfying $\lim_{n\to\infty}na_n=0$. Then, is it true that $\sum_{n=1}^\infty a_n<\infty$?

I think the answer is yes from my intuition since $a_n$ decreases faster than $1/n$ as $n\to\infty$, i.e. $$\exists \alpha>0\ s.t.\ a_n\sim O(1/n^{1+\alpha}),$$ and we know that $\sum_{n=1}^\infty 1/n^s$ converges if $s>1$. Actually this is the converse of this problem.

How can I prove this? Thank you in advance.

It is not true. $a_n = \frac{1}{n\ln(n)}$. – User8128 Jul 27 '17 at 02:17 — User8128, Jul 27 '17 at 02:17

Ben Grossmann · Accepted Answer · 2017-07-27T02:21:26.223

8

The statement is false. For example, you can verify that $$ \sum_{n=2}^\infty \frac{1}{n \ln(n)} $$ satisfies your criterion but fails to converge (which you can verify by the integral test).

More generally: if $f: \Bbb R \to \Bbb R$ is any increasing function with $\lim_{n \to \infty} f(x) = \infty$, we'll find that $a_n = f(n) - f(n-1)$ leads to a divergent series (no matter how slowly $f$ diverges).

edited Jul 27 '17 at 02:21

answered Jul 27 '17 at 02:18

Ben Grossmann

225,327

As @user8128 indicated in the comments – gen-ℤ ready to perish Jul 27 '17 at 02:19
Okay, wonderful. – gen-ℤ ready to perish Jul 27 '17 at 02:22
Oh, very simple counterexample! Thank you both. – Tom TJ Jul 27 '17 at 02:52

$\lim_{n\to\infty}na_n=0\Rightarrow\sum_{n=1}^\infty a_n<\infty$?

1 Answers1