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

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.

• Do you know the AM-GM inequality? Commented Jan 22, 2014 at 8:14
• The title and the question do not match. Please edit. Commented Jan 22, 2014 at 8:18
• How AM GM will work Erick Wong? can you explain? Commented Jan 22, 2014 at 8:19
• Obviously the $a_n$ cannot be positive integers. Commented Jan 22, 2014 at 8:31
• The statement is vacuously true since $\sum n^2a_n^2 =\infty$. Commented Jan 22, 2014 at 8:37

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 $\sum a_n < \infty$

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 $a_n = 0$)

• Take $a_n=\frac{1}{n^{\frac32}}$ : $\sum a_n$ converges, yet $\sum n^2 a_n^2=\sum \frac1n$ diverges... Commented Jan 22, 2014 at 8:40
• @OlivierBégassat Given that $a_n$ positive Commented Jan 22, 2014 at 8:44
• The counter example I gave is positive. Commented Jan 22, 2014 at 8:46
• @OlivierBégassat I'm sorry, I meant positive integers. your example is not Integers Commented Jan 22, 2014 at 8:48
• I see, the OP indeed considers an integer sequence... I'm not quite sure why since he or she accepted goaxinge's answer which is the proof of the general case. Commented Jan 22, 2014 at 8:48