Sum of Infinite Geometric Progression

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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 $\left \vert {z}\right \vert < 1$.

So by Sequence of Powers of 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.


It remains to demonstrate absolute convergence:

The absolute value of $\left \vert {z}\right \vert$ is just $\left \vert {z}\right \vert$.

By assumption:

$\left \vert {z}\right \vert < 1$

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


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


Proof 2

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

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

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.


Proof 3

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


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

Hence the result.