How to prove or disprove - If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty na_n$ both converge, then $\lim_{n\to\infty}n^2a_n=0$

Question

My question is: whether the following is true or not, and how to prove in either case:

$$\text{If}\quad\sum_{n=1}^\infty a_n\quad\text{and}\quad\sum_{n=1}^\infty na_n\quad\text{both converge, then}\quad \lim_{n\to\infty}n^2a_n=0$$

There's no additional condition like $a_n$'s are positive, etc. (i.e. $a_n$ could be either positive or negative.)

(*) As a background, this question came to when I was solving this problem: $$\text{If}\quad\sum_{n=1}^\infty a_n=1\quad\text{and}\quad\sum_{n=1}^\infty na_n=2\text{, then}\quad \lim_{n\to\infty}n^2(a_n-a_{n+1})=?$$ that leads to estimate $\sum_{n=1}^\infty (n^2a_n-(n+1)^2a_{n+1})$ and hence $\lim_{n\to\infty}n^2a_n$.

How about reducing it to $$\sum_{n=1}^\infty b_n\implies \lim_{n\to\infty}nb_n=0$$where $b_n=na_n$ — Simply Beautiful Art, Sep 07 '17 at 20:54
Prove that if $\sum_{n=1}^ \infty na_n$ converges, then $\sum_{n=1}^ \infty a_n$ converges. and Series converges implies $\lim na_n=0$ — Simply Beautiful Art, Sep 07 '17 at 20:56
@SimplyBeautifulArt, your first comment, and the second link of your second comment, does not apply without additional hypotheses on ${b_n}$, like positive and nonincreasing. For example, consider $b_n=\begin{cases} 1/n & \exists k\in \mathbb{N}, n=10^k\ 0 & \text{otherwise}\end{cases}$. — vadim123, Sep 07 '17 at 21:00
For the background question, I suppose that $a_n$ are positive? — clark, Sep 07 '17 at 21:53

score 5 · Accepted Answer · answered Sep 07 '17 at 20:56

5

Consider the sequence $(a_n)_n$ defined by $$a_n = \frac{(-1)^n}{n^2}$$ for $n\geq 1$.

answered Sep 07 '17 at 20:56

Clement C.

67,323

1

:-) Yup, likely it is wanted that the terms are all positive, but idk... – Simply Beautiful Art Sep 07 '17 at 20:58
@SimplyBeautifulArt Not really: "There's no additional condition like $a_n$'s are positive, etc. (i.e. $a_n$ could be either positive or negative.)" – Clement C. Sep 07 '17 at 21:02
Hence the "idk..." part, seeing as it was explicitly stated in the original post. – Simply Beautiful Art Sep 07 '17 at 21:13
One probably needs positive and monotone for the proposed limit to hold now that I think about it. – Simply Beautiful Art Sep 07 '17 at 21:16
Thanks. This seems to be the best example. – Kay K. Sep 07 '17 at 21:23
@KayK. Probably the simplest, but mechanodroid's is worth keeping in mind as well to see that non-negativity is not sufficient for the conclusion you wanted to hold. – Clement C. Sep 07 '17 at 21:24
Yeah, I can see it now. Thanks. – Kay K. Sep 07 '17 at 21:30

score 4 · Answer 2 · answered Sep 07 '17 at 21:11

For an example where $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty na_n$ converge absolutely define the sequence $a_n$ as:

$$a_n = \begin{cases} \frac{1}{n^2}, & \text{if $n$ is a perfect square} \\ 0, & \text{otherwise} \end{cases}$$

We have:

$$\sum_{n=1}^\infty a_n = \sum_{k=1}^\infty \frac{1}{k^4} < +\infty$$ $$\sum_{n=1}^\infty n \cdot a_n = \sum_{k=1}^\infty k^2\cdot\frac{1}{k^4} = \sum_{k=1}^\infty \frac{1}{k^2} < +\infty$$

However,

$$n^2a_n = \begin{cases} 1, & \text{if $n$ is a perfect square} \\ 0, & \text{otherwise} \end{cases}$$

so it does not converge.

hamam_Abdallah · Answer 3 · 2017-09-07T21:01:38.053

1

Take $$a_n=\frac {\sin (n)}{n^2} $$

$$\sum a_n $$ and $$\sum na_n $$ are convergent by Dirichlet's test , but $$\lim n^2a_n \ne 0$$

edited Sep 07 '17 at 21:01

answered Sep 07 '17 at 20:59

hamam_Abdallah

62,951

=P Late to the party – Simply Beautiful Art Sep 07 '17 at 20:59

How to prove or disprove - If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty na_n$ both converge, then $\lim_{n\to\infty}n^2a_n=0$

3 Answers3