Sum of Infinite Geometric Sequence/Proof 2

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Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.

Let $z \in S$.

Let $\size z < 1$, where $\size z$ denotes:

the absolute value of $z$, for real and rational $z$
the complex modulus of $z$ for complex $z$.

Then $\ds \sum_{n \mathop = 0}^\infty z^n$ converges absolutely to $\dfrac 1 {1 - z}$.


By the Chain Rule for Derivatives and the corollary to $n$th Derivative of Reciprocal of $m$th Power:

$\dfrac {\d^n} {\d z^n} \dfrac 1 {1 - z} = \dfrac {n!} {\paren {1 - z}^{n + 1} }$

Thus the Maclaurin series expansion of $\dfrac 1 {1 - z}$ is:

$\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} \dfrac {n!} {\paren {1 - 0}^{n + 1} }$

whence the result.