Sum of Infinite Geometric Progression

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

$\Box$


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 $z$.

Hence:

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

$\blacksquare$


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.

$\blacksquare$


Sources