Determine whether the infinite series $sum_{n=1}^{infty}frac{n!}{n^3}$ is convergent

The question asks to use the direct comparison test to determine whether

$$\sum_{n=1}^{\infty}\frac{n!}{n^3}$$

is convergent or divergent.

I was wondering whether the direct comparison test requires that a series consist of a positive sequence, as the only thing I could think of to compare the series to was:

$$\sum_{n=1}^{\infty}-\frac{n^2}{n^3} \leq \sum_{n=1}^{\infty}\frac{n!}{n^3}$$

with the LHS being a negative harmonic series that diverges and hence shows the series on the right diverges as well.

Is this reasoning correct/is there an easier way to do this?

Answer

Notice that $n! > n^3$ for all sufficiently large $n$, and

$$\sum_{n = 1}^{\infty} 1 = \infty$$