Proving an alternating infinite series to be divergent I am trying to prove $$\sum_{n=0}^\infty \frac{(-4)^{3n}}{5^{n-1}}$$ is a divergent series. I know that it increases as the series progresses regardless of sign, so it must be divergent, but im not sure how to prove it. 
Usually with an alternating infinite series as long as the term not involving the minus doesnt tend to zero then it is easy to prove, however in this, that term is $$\frac{1}{5^n-1}$$ (I think), so it makes it a little harder, maybe i'm taking the wrong appraoch to this,  any help?

Hint:
$$\dfrac{(-4)^{3n}}{5^{n-1}}=5\left(\dfrac{-64}5\right)^n$$
Use this
