Series with $\sum a_n$ converges but $\sum n a_n^2$ diverges, and $a_n$ is decreasing

The answer is no, there are two steps to do:

1) If $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, show that $a_n \leq 1/n$ for all sufficiently large $n$.

2) Conclude that if $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, then $\sum_{n=1}^{\infty} na_n^2 < \infty$.

The part 1 of @Michael's answer lies on the following fact: $na_{n}\rightarrow 0$ for decreasing/nonincreasing $(a_{n})$, $a_{n}\geq 0$ and that $\displaystyle\sum a_{n}<\infty$.

Consider $na_{2n}\leq a_{2n}+a_{2n-1}+\cdots+ a_{n+1}$ and also consider $na_{2n+1}$. Finally, consider $(2n+1)a_{2n+1}=2(na_{2n+1})+a_{2n+1}$. Note that we have $a_{n}\rightarrow 0$. So now consider the even subsequence and odd subsequence of $(ka_{k})$.