Sum of Infinite Geometric Sequence

Theorem
Let $S$ be a standard number field, i.e. $\Q$, $\R$ or $\C$.

Let $z \in S$.

Let $\left \vert {z}\right \vert < 1$, where $\left \vert {z}\right \vert$ denotes:
 * the absolute value of $z$, for real and rational $z$
 * the complex modulus of $z$ for complex $z$.

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

Corollary
With the same restriction on $z \in S$:


 * $\displaystyle \sum_{n=1}^\infty z^n = \frac z {1-z}$

Proof
From Sum of Geometric Progression, we have:
 * $\displaystyle s_N = \sum_{n=0}^N z^n = \frac {1 - z^{N+1}} {1 - z}$

We have that $\left \vert {z}\right \vert < 1$.

So by Power of a Number Less Than One:
 * $z^{N+1} \to 0$ as $N \to \infty$

Hence $s_N \to \dfrac 1 {1 - z}$ as $N \to \infty$.

The result follows.

To demonstrate absolute convergence we note that
 * $\displaystyle \sum_{n=0}^\infty \left \vert {z}\right \vert^n = \frac 1 {1-\left \vert {z}\right \vert}$

as $\left \vert {z}\right \vert$ fulfils the same condition for convergence as $z$.

Proof of Corollary
Or: