# Sum of Infinite Geometric Sequence

## Theorem

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

### Corollary 1

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

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

### 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 Sequence, we have:

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

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:

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

$\blacksquare$

## Proof 2

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

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.

$\blacksquare$

## Proof 3

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

Then:

 $\ds z S$ $=$ $\ds z \sum_{n \mathop = 0}^\infty z^n$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty z^{n + 1}$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty z^n$ Translation of Index Variable of Summation $\ds$ $=$ $\ds S - z^0$ $\ds$ $=$ $\ds S - 1$ $\ds \leadsto \ \$ $\ds 1$ $=$ $\ds \paren {1 - z} S$

Hence the result.

$\blacksquare$