Sum of Infinite Geometric Sequence/Proof 1

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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 $\dfrac 1 {1 - z}$.


Proof

From Sum of Geometric Sequence, we have:

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

We have that $\size z < 1$.

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.

$\Box$


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:

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

$\blacksquare$


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