Sum of Infinite Geometric Progression/Proof 1

From ProofWiki
Jump to: navigation, search


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 $\dfrac 1 {1 - z}$.


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 $z$.


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