# The series $\sum a_n$ is conditionally convergent. Prove that the series $\sum n^2 a_n$ is divergent.

Ratio and root tests won't help. And I can't use the comparison test because $|a_n|$ is not necessarily smaller than $n^2a_n$.

Can I use limits? We know:

1. $\lim\limits_{n \to \infty} a_n = 0$
2. $\lim\limits_{n \to \infty} |a_n| \ne 0$

And we need to prove:

1. $\lim\limits_{n \to \infty} n^2a_n \ne 0$

Any ideas/hints?

• Your second point is incorrect: since $\lim a_n=0$, we have $\lim|a_n|=0$ too. And we don't have to prove $\lim n^2a_n\ne0$: this would be sufficient but it is not necessary. – David Mar 20 '14 at 1:31

Suppose that $n^2 a_n$ had a convergent series: That is,

$$\sum_{n = 1}^{\infty} n^2 a_n = \ell \in \mathbb{R}$$

Then $n^2 a_n \to 0$ as $n \to \infty$, so $n^2 a_n$ is eventually less than $1$ in absolute value, implying that $$|a_n| < \frac 1 {n^2}$$

for sufficiently large $n$. Now think about the comparison test.

• Then I take the limit of both sides. The limit of $|a_n|$ =/=. Since we're taking the absolute value, it will be greater than 0. The limit of $1/n^2$ = 0. LS greater than 0 cannot be less than 0, so there is a contradiction. Correct? – Bob Mar 20 '14 at 1:54
• Just knowing that the limit of $1/n^2$ is zero isn't enough (consider the limit $1/n$, for example). Just compare directly to the series $\sum_{n = 1}^{\infty} 1/n^2$. – user61527 Mar 20 '14 at 1:57
• What am I comparing to the series $1/n^2$? The sum of the series $n^2a_n$ ? But if it's less than $1/n^2$, and $1/n^2$ is convergent, that tells me $n^2a_n$ is convergent, which is the wrong answer. I'm very confused. – Bob Mar 20 '14 at 2:09
• If we were to suppose that $\sum n^2 a_n$ is convergent, then compare $\sum |a_n|$ to $\sum 1/n^2$. From the above scratchwork, we'll find that $\sum |a_n|$ converges, which is quite bad. – user61527 Mar 20 '14 at 2:23
• Ohhhh I see. Thank you. – Bob Mar 20 '14 at 2:41