Proof of divergence

Question

I am genuinely sorry but I have a seemingly very easy proof to do and I can't see at all where to even start:

$u_n = u_0q^n$ with $u_0 \neq 0$ and $q < -1$.

I just need to show that $(u_n)$ is divergent.

Thank you!

Answer 1

It is divergent because: $$\lim_\limits{n\to +\infty} u_{2n}=\text{sgn}(u_0)\infty \ne \lim_\limits{n\to +\infty} u_{2n-1}=-\text{sign}(u_0)\infty$$

Answer 2

We have $|q|>1$, hence $|q|=1+h$, with some $h>0$. It follows by Bernoulli:

$|q|^n \ge 1+nh >nh$.

Hence $|u_n| >|u_0|nh$ for all $n$. The sequence $(u_n)$ is therefore unbounded and hence divergent.