$\displaystyle \sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}$
the dividing of a_n+1 /a_n is not conlusive
?
Now, you have me confused! $\displaystyle a_{n+1}- a_n< 0$ is the same as $\displaystyle a_{n+1}< a_n$ and that is the definition of "decreasing".
Or are you saying that you know you have to prove one of those but don't know how?
I would look at $\displaystyle \frac{a_{k+1}}{a_k}= \frac{(k+1)^2}{e^{(k+1)^3}}\frac{e^{k^3}}{k^2}$$\displaystyle = \frac{(k+1)^2}{k^2}e^{k^3-((k+1)^3}= \frac{(k+1)^2}{k^2}e^{-3k^2- 3k- 1}$.
Now, in the limit, as k goes to infinity, that fraction goes to 1 so this "eventually" decreases to 0.
If you really need to prove that it is decreasing for all k greater than or equal to 1, I recommend Plato's suggestion. Take the dervative of $\displaystyle x^2e^{-x^3}$.