$$\forall a,b\in\mathbb R:\lvert b\rvert\gt1\rightarrow\sum_{n=1}^∞\frac{a}{b^n}=\frac{a}{(\lvert b\rvert-1)\frac{1}{b}\lvert b\rvert}$$
If $b$ is restricted to be non-negative, the expression simplifies to $$\forall a,b\in\mathbb R:b\gt1\rightarrow\sum_{n=1}^∞\frac{a}{b^n}=\frac{a}{b-1}$$ The denominator $(\lvert b\rvert-1)\frac{1}{b}\lvert b\rvert$ from the right-hand side of the identity in the first expression is the only way I could come up with that preserves $b$'s sign when it is negative. This is the part I'm trying to simplify.
Thanks in advance for any help.
Using the standard geometric series $$\sum_{n=0}^\infty x^n=\frac1{1-x}\qquad (|x|\lt1)$$ We have that your result is just $$\frac{a}b\sum_{n=0}^\infty(1/b)^n=\frac{\frac{a}b}{1-\frac1{b}}=\frac{a}{b-1}\qquad(|b|\gt1)$$which is also valid for negative $b$.