# Sum of Infinite Geometric Progression

## Theorem

Let $S$ be a standard number field, that is $\Q$, $\R$ 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 $\displaystyle \sum_{n \mathop = 0}^\infty z^n$ converges absolutely to $\dfrac 1 {1 - z}$.

### Corollary 1

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

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

### Corollary 2

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

$\displaystyle \sum_{n \mathop = 0}^\infty a z^n = \frac a {1 - z}$

## Proof 1

From Sum of Geometric Progression, we have:

$\displaystyle s_N = \sum_{n \mathop = 0}^N z^n = \frac {1 - z^{N + 1} } {1 - z}$

We have that $\size z < 1$.

$z^{N + 1} \to 0$ as $N \to \infty$

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

The result follows.

$\Box$

It remains to demonstrate absolute convergence:

The absolute value of $\size z$ is just $\size z$.

By assumption:

$\size z < 1$

So $\size z$ fulfils the same condition for convergence as $z$.

Hence:

$\displaystyle \sum_{n \mathop = 0}^\infty \size z^n = \frac 1 {1 - \size z}$

$\blacksquare$

## Proof 2

$\dfrac {\mathrm d^n}{\mathrm dz^n} \dfrac 1 {1 - z} = \dfrac {n!} {\left({1 - z}\right)^{n + 1}}$

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

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} \dfrac {n!} {\left({1 - 0}\right)^{n + 1}}$

whence the result.

$\blacksquare$

## Proof 3

Let $S = \displaystyle \sum_{n \mathop = 0}^\infty z^n$.

Then:

 $\displaystyle z S$ $=$ $\displaystyle z \sum_{n \mathop = 0}^\infty z^n$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty z^{n + 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty z^n$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle S - z^0$ $\displaystyle$ $=$ $\displaystyle S - 1$ $\displaystyle \leadsto \ \$ $\displaystyle 1$ $=$ $\displaystyle \left({1 - z}\right) S$

Hence the result.

$\blacksquare$