Intuitively my guess is that the statement is true but i've struggled to find a way to show it rigorously. The reason I believe it to be true is since $\sum n^2 a_n$ converging seems like a "strong" condition and for some obvious candidates that satisfy this (e.g. $\sum \frac{1}{n^4}$), $\sum n^2 a_n$ converging usually meant that that $\sum a_n$ converges absolutely.
This line of thought led me to attempting to prove the contrapositive instead; that if $\sum n^2a_n$ converges, then $\sum a_n$ converges absolutely (and thus not conditionally).
I've tried to show this using the Cauchy criterion, which eventually brought me to attempting to prove the following inequality:
$$|a_{n+1}| + |a_{n+2}| + \cdots + |a_m| \le |(n+1)^2a_{n+1} + \cdots + m^2a_m|$$
For $m > n \ge N$ for some $N \in \mathbb{N}$ but i've struggled to make meaningful progress past this point. My main issue with the inequality above being that I don't have any guarantee that each of the terms $a_n$ have the same sign, which made it difficult to work with the expression on the right.
Is there something fundamentally wrong with my line of thought above? Any hints on how I might better approach the problem would be really appreciated.
Best Answer
If $\sum n^{2}a_n$ converges then $n^{2}a_n \to 0$. In particular $n^{2}a_n$ is bounded. Hence $|a_n| \leq \frac C {n^{2}}$ for some finite constant $C$ which makes $\sum a_n$ absolutely convergent.