$$\sum_{n=1}^\infty\ e^{an}\,n^2 \quad\text{for}\;\;a\in R $$
**Hint** : Use the **_root test_** :
To determine whether $\;\;\sum_{n=1}^\infty b_n\;\;$converges or diverges, evaluate $\;\;\lim_{n\to \infty}\sqrt[\large n]{|b_n|}.\;\;$ In your series, $\;b_n > 0 \;\;\forall n,\;$ so we can drop the absolute value sign:
$$\text{We use the fact that:}\;\; \lim_{n\to \infty}\sqrt[\large n]{n^2} = 1,$$
$$\lim_{n\to \infty} \sqrt[\large n]{e^{an}n^2} \;=\; \lim_{n\to \infty} \sqrt[\large n]{e^{an}}\cdot \sqrt[\large n]{n^2} \;=\; \lim_{n\to \infty}\sqrt[\large n]{e^{an}}\; = \;e^a$$
* For what $\;a\;$ is $\;e^a < 1\;$? (At those values, the given series converges.)
* For what values of $\;a\;$ is $\; e^a \gt 1\;$? (At those values, the series diverges.)