Solution
= Solution
For $a_n=n^2e^{-\alpha n}$, the <ratio test> gives
$$
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\left(1+\frac1n\right)^2e^{-\alpha}=e^{-\alpha}.
$$
This is less than one exactly when $\alpha>0$. For $\alpha=0$ the terms are $n^2$, and for $\alpha<0$ they grow exponentially; in both cases they fail to tend to zero. Consequently
$$
\boxed{\text{convergent exactly for }\alpha>0.}
$$