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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 |

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.

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    I'm so sorry I made a mistake of distraction. – 2017-02-26
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    Your conclusion is true if $b$ is positive. We tend to not use the notation $\sqrt[b]{a}$ when $b<0$, but if you wanted to include that case, it is not true then. – 2017-02-26

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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 $$\left| \frac{z^b}{a} \right| < 1 \qquad \Longleftrightarrow \qquad |z| < \sqrt[b]{a}.$$

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    Only as long as `a > 0`. `a < 0` implies `z^b > 1` – 2017-02-26
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    Of course is $a>0$. – 2017-02-26