Sum of Infinite Geometric Sequence

Theorem
Let $z \in \C$ be a complex number.

Let $\left\vert z \right\vert < 1$. Here $\left\vert z\right\vert$ denotes the modulus (also called absolute value) of $z$.

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

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


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

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

If $\left\vert z \right\vert < 1$, then 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: