"fast enough" decay of an $\ell ^2$ sequence implies $\ell ^1$? To be specific, say we are given that $(a_n)$ is a sequence of real numbers such that \begin{equation} \sum_{n=1}^{\infty} n^3 a_n^2 < \infty. \end{equation}
Is it then true that 
$$ \sum_{n=1}^{\infty} a_n < \infty ?$$
I believe the answer is yes, because it seems that $$\sum_{n=1}^{\infty} n^3 a_n^2 < \infty$$ implies that $a_n^2 \le O(\frac{1}{n^4})$ and therefore $$ \sum_{n=1}^{\infty} a_n \le \sum_{n=1}^{\infty} \frac{1}{n^4} < \infty. $$
Is this intuition correct? If so, I'm having trouble formalizing the reason why it must be true that $a_n^2 \le O(\frac{1}{n^4})$, so providing a formal proof would be helpful, preferably using concepts no more advanced than you'd see in the first semester of an applied analysis course.
 A: By the Cauchy-Schwarz inequality,
$$ \left(\sum_{n\geq 1}n^3 a_n^2\right)\cdot\left( \sum_{n\geq 1}\frac{1}{n^3}\right) \geq \left(\sum_{n\geq 1}|a_n|\right)^2 \tag{1}$$
hence:
$$ \sum_{n\geq 1}|a_n| \leq \sqrt{\zeta(3)\cdot\sum_{n\geq 1}n^3 a_n^2}.\tag{2}$$
A: The sum $\sum n^3 a_n^2$ converges, and therefore $\lim_{n\to\infty} n^3a_n^2=0$. So after a while the terms $n^3a_n^2$ are $\lt 1$.  So after a while $|a_n|\lt \frac{1}{n^{3/2}}$. By the Comparison Test we have absolute convergence, and therefore convergence.
A: One proof uses only that $n^3 a_n^2\le c $  bounded and and gets $\sum|a_n| \le c \zeta(\frac{3}{2})$, the other uses Cauchy-Schwarz and gets $\sum|a_n| \le C \zeta(3)^{\frac{1}{2}}$. However, the second proof can get away with only the convergence of $\sum n^{1+\epsilon} a_n^2$. In general, if:
$$\sum n^{p\alpha} |a_n|^{p}< \infty$$
using the Hölder's inequality we get:
$$\sum n^{p \alpha } |a_n|^{p} \cdot \sum n^{-q \alpha } \ge \sum |a_n|  $$
and so we get the convergence of $\sum |a_n|$ if $q \alpha = \frac{p \alpha}{p-1} >1$. However, only using that $n^{p \alpha} |a_n|^p$ is bounded would imply convergence if we knew that $\alpha>1$, a stronger condition on $\alpha$. 
