Let $b \in [0,1)$. Prove that $\frac{b}{1-b} \in [0,\infty)$

Another method is to note that \begin{equation} \dfrac{b}{1-b}=\sum_{i=1}^\infty b^i \end{equation} Since each term is non-negative, therefore, sum of this series is also non-negative.

Define the function $ f $ from $[0,1)$ to $ \Bbb R$ by

$$(\forall x\in[0,1))\;\; f(x)=\frac{x}{1-x}$$

$ f $ is continuous at $ [0,1)$.

$ f $ is differentiable at $ [0,1)$ and

$$(\forall x\in[0,1))\;\; f'(x)=\frac{1-x+x}{(1-x)^2}>0$$ $ f $ is then strictly increasing at $ [0,1)$.

Thus, $ f $ is a bijection from $ [0,1)$ to $$f([0,1))=[f(0),\lim_{x\to 1^-}f(x))=[0,+\infty)$$

we conclude that $$(\forall b\in[0,1))\;\; f(b)=\frac{b}{1-b}\ge 0$$

Remark:

You can simply say $$0\le b<1\; \implies$$ $$b\ge 0 \text{ and } 1-b>0 \;\implies$$ $$\frac{b}{1-b}\ge 0\; \implies$$ $$\frac{b}{1-b}\in [0,+\infty)$$