Does the series $ \sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$ converge or diverge? Does the series $$\sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$$ converge or diverge?
I have attempted using the alternating series test, but couldn't find a useful function to use as $b_n$ and no other tests I know seem to be useful in coming to the conclusion of whether it is convergent or divergent.

The series converges absolutely. This is because we have $$0 \leq \exp\left(-\frac{n^3}{3}\right) \leq \frac{1}{1 + \frac{n^3}{3} + \frac{n^6}{18}}$$
by considering the Taylor expansion  and hence 
$$0 \leq n^2\exp\left(-\frac{n^3}{3}\right) \leq \frac{n^2}{1 + \frac{n^3}{3} + \frac{n^6}{18}} \leq 18n^{-4}$$
and $$\sum 18n^{-4}$$ converges by the integral test. 
More intuitively, just keep in mind that exponentials of the form $a^{-x}$ for $a>1$ decay faster than any rational function. 
If you want to use the alternating series test to merely establish conditional convergence, you can use L'hopital to show $\lim_{x \to \infty} x^2e^{-\frac{x^3}{3}} = 0$
