If $\sum n^2 a_n^2<\infty$ then $\sum a_n$ is convergent. [duplicate]

Question

Let $\{a_n\}$ be a sequence of positive integers, then prove/disprove:

If $\sum n^2 a_n^2<\infty$ then $\sum a_n$ is convergent.

Answer 1

We can use the Cauchy—Schwarz Inequality in $l^\infty$.

$(\sum |a_n||b_n||)^2 \leq (\sum a_n^2) \times (\sum b_n^2)$

Then we have

$(\sum |a_n|)^2 \leq (\sum n^2 a_n^2) \times (\sum \frac{1}{n^2})$

Because

$\sum n^2 a_n^2 < \infty$ and $\sum \frac{1}{n^2} < \infty$

So $\sum a_n < \infty$

Answer 2

You can use the Direct comparison test

You can easily see that $ \sum n^2 a_n^2 > \sum a_n$.

By the comparison test, you can prove the the series $\sum a_n$ converge $\iff \sum n^2 a_n^2$ does.

Even thought it would never converge since $\lim a_n \ne 0$ (unless $a_n = 0$)