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Develop $$ \sum_{n=1}^{\infty} z^n $$ in a Taylor series, around $z=a$ with $|a|<1.$ What is the new radius of convergence?

Solution. Now i think this is a "clitche", but i know that $$\frac{1}{1-z}=\sum_{n=1}^{\infty} z^n $$ for $|z|<1.$ Now if i take $$ \frac{1}{1-z} = \frac{1}{1-a} \, \frac{1}{1-\frac{z-a}{1-a}} = \frac{1}{1-a} \, \sum_{n=1}^{\infty} \left(\frac{z-a}{1-a}\right)^n $$ for $|\frac{z-a}{1-a}|<1$, so $|z-a|<|1-a|$ is the new radius of convergence. Is it right? Cause i'm a little confused. Is this the representation of an analytic function in that disk? Is the analytic continuation of the original function? i had never learn this in a simple words. Thanks for your time.

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    Almost. The series begins at $n=0$ – 2017-02-21
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    @Dr.MV How is that affect my result? – 2017-02-21
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    It affects the result in that the series as written, does not represent $(1-z)^{-1}$. Other than this mistake, everything else is fine. – 2017-02-21
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    @Dr.MV so, replace $\frac{1}{1-z}$ for $\frac{z}{1-z}$ and the rest still remains (with that little adjust) – 2017-02-21
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    Yes, or replace $n=1$ with $n=0$ – 2017-02-21

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The power series $f(z)=\sum_{n=0}^\infty z^n$ defines an analytic function in the disk $D(0,1)$ which extends to $\mathbb C\setminus\{1\} $. Therefore, this function can be expressed as a power series around any $a\in\mathbb C\setminus\{1\}$, and its radius of convergence is exactly the distance of $a$ from $1$. Indeed $$ f(z)=\frac{1}{1-a}\sum_{n=0}^\infty\left(\frac{z-a}{1-a}\right)^n\!, $$ for every $z\in D(a,|1-a|)$.