Proving an alternating infinite series to be divergent

Question

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?

Answer 1

Since $\lim_{n\to\infty}\left\lvert\frac{(-4)^{3n}}{5^{n-1}}\right\rvert=\infty$, you don't have $\lim_{n\to\infty}\frac{(-4)^{3n}}{5^{n-1}}=0$, and therefore the series diverges.

Answer 2

Hint:

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

Use this

Answer 3

Hint

Use the d'alembert Ratio Test.