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.

  • $\begingroup$ Do you know the AM-GM inequality? $\endgroup$
    – Erick Wong
    Jan 22, 2014 at 8:14
  • $\begingroup$ The title and the question do not match. Please edit. $\endgroup$ Jan 22, 2014 at 8:18
  • $\begingroup$ How AM GM will work Erick Wong? can you explain? $\endgroup$
    – user121418
    Jan 22, 2014 at 8:19
  • 3
    $\begingroup$ Obviously the $a_n$ cannot be positive integers. $\endgroup$ Jan 22, 2014 at 8:31
  • $\begingroup$ The statement is vacuously true since $\sum n^2a_n^2 =\infty$. $\endgroup$
    – pritam
    Jan 22, 2014 at 8:37

2 Answers 2


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

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


$\sum n^2 a_n^2 < \infty$ 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 $a_n = 0$)

  • $\begingroup$ Take $a_n=\frac{1}{n^{\frac32}}$ : $\sum a_n$ converges, yet $\sum n^2 a_n^2=\sum \frac1n$ diverges... $\endgroup$ Jan 22, 2014 at 8:40
  • $\begingroup$ @OlivierBégassat Given that $a_n$ positive $\endgroup$
    – Billie
    Jan 22, 2014 at 8:44
  • $\begingroup$ The counter example I gave is positive. $\endgroup$ Jan 22, 2014 at 8:46
  • $\begingroup$ @OlivierBégassat I'm sorry, I meant positive integers. your example is not Integers $\endgroup$
    – Billie
    Jan 22, 2014 at 8:48
  • $\begingroup$ 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. $\endgroup$ Jan 22, 2014 at 8:48

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