I've already proved that for any $p > 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.
Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| < 1$.
I've split it into two possible cases:
(1) $0 < x < 1$. Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p > 0$ (since $x < 1$). Then we let $\alpha = 0$ to see that $\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$
(2) $-1 < x < 0$. Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 < 0$).
How can I prove Case (2)?