$\frac{1}{1-\frac{z^{b}}{a}}=\sum \left ( \frac{z^{b}}{a} \right )^{n} \; \; \; \; \; \; \left | z \right |<\sqrt[b]{a}$

Question

I have a doubt on Taylor's series. I always used the formula:

$$\frac{1}{1-\frac{z}{a}}=\sum \left ( \frac{z}{a} \right )^{n} \; \; \; \; \; \; \left | z \right |<a$$

Is it always true also the following

$$\frac{1}{1-\frac{z^{b}}{a}}=\sum \left ( \frac{z^{b}}{a} \right )^{n} \; \; \; \; \; \; \left | z \right |<\sqrt[b]{a}$$

Thank you so much.

Answer 1

Short answer: $$\frac{1}{1-\frac{z^{b}}{a}}=\sum_{n=0}^\infty \left ( \frac{z^{b}}{a} \right )^{n} \; \; \; \; \; \; \left | z \right |<\sqrt[b]{a}$$ holds for positive $b\in \mathbf R$ since we have $$\left| \frac{z^b}{a} \right| < 1 \qquad \Longleftrightarrow \qquad |z| < \sqrt[b]{a}.$$