$\{a_n\}$ is a bounded sequence, $|q| < 1$. Prove that $a_0 + a_1 q + a_2 q^2 + \cdots + a_n q^n$ is a Cauchy sequence

$$\{a_n\}$$ is a bounded sequence (so it has upper and lower bound), $$|q| < 1$$. Prove that $$s_n = a_0 + a_1 q + a_2 q^2 + \cdots + a_n q^n$$ is a fundamental sequence (Cauchy sequence).

What I'll be writing next is not a proof, it's just a mere assessment, so I can show my "thinking process".
Well, let $$b = \max \{a_n\}$$. Then $$a_0 + a_1 q + a_2 q^2 + \cdots + a_n q^n \leq b + bq + bq^2 + \cdots + bq^n = \frac{b(1-q^n)}{1-q}$$ $$q^n \to 0, n \to \infty$$, so $$\frac{b(1-q^n)}{1-q} \to \frac{b}{1-q}$$ So we've got some fixed number $$\frac{b}{1-q}$$ (since $$b$$ and $$q$$ are also fixed numbers). At least, according to my assumption, it doesn't "converge" to infinity.
With $$\varepsilon$$-notation the sequence is fundamental, if $$\forall \varepsilon$$ $$\exists N \in\mathbb{N}$$ $$\forall n \geq N$$ $$\forall p \geq 1:$$ $$|a_{n+p} - a_n| < \varepsilon$$
Then $$|s_{n+p} - s_n| \leq a_0 + a_1 q + a_2 q^2 + \cdots + a_{n+1} q^{n+1} + a_{n+2} q^{n+2} + \cdots + a_{n+p} q^{n+p} - a_0 - a_1 q - a_2 q^2 - \cdots - a_n q^n$$ $$|s_{n+p} - s_n| \leq a_{n+1} q^{n+1} + a_{n+2} q^{n+2} + \cdots + a_{n+p} q^{n+p}$$ Obviously, I can make the same move with $$b = \max \{a_n\}$$, but then: $$|s_{n+p} - s_n|\leq \frac{b(1-q^{n+p})}{1-q} - \frac{b(1-q^n)}{1-q} = \frac{b}{1-q}(q^n - q^{n+p})$$ Further we need to "extract" $$n$$ from expression $$\frac{b}{1-q}(q^n - q^{n+p}) < \varepsilon$$ in order to find $$N$$ for " $$\forall \varepsilon$$ $$\exists N \in\mathbb{N}$$ $$\forall n \geq N$$ $$\forall p \geq 1:$$ $$|a_{n+p} - a_n| < \varepsilon$$ "
And while I'm sure that $$1-q > 0$$, I have no reason to assume that $$b > 0$$, so inequality sign is unknown in case of division both sides by $$\frac{b}{1-q}$$ Hence, something is wrong.

It's not too hard to remedy this. All you have to do is be more careful with taking absolute values - note $$q$$ isn't necessarily positive either. Below, I treat $$b$$ as $$\max|a_n|$$, so that it is positive.
Fix $$n,m\in\Bbb N$$. A difference of partial sums: \begin{align}|s_{n+m}-s_n|&=|a_0+\cdots+a_{n+m}q^{n+m}-a_0-\cdots-a_nq^n|\\&\le|a_{n+1}q^{n+1}+\cdots+a_{n+m}q^{n+m}|\\&\le b\cdot\sum_{j=1}^m|q|^{n+j}\\&=b\cdot\frac{1-|q|^m}{1-|q|}\cdot|q|^{n+1}\\&<\frac{b}{1-|q|}\cdot|q^n|\end{align}
Now this last term is bounded independently of $$m$$, and as $$|q|<1$$ for all $$\epsilon>0$$ there will exist $$N\in\Bbb N$$ that $$\frac{b}{1-|q|}|q|^n<\epsilon$$ for all $$n\ge N$$. You can explicitly take: $$N:=\left\lceil\frac{1}{\log|q|}\log\left(\frac{\epsilon}{b}(1-|q|)\right)\right\rceil$$Where it is important to note that $$\log|q|$$ is negative.