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 $\ds \sum_{n \mathop = 0}^\infty z^n$ converges absolutely to $\dfrac 1 {1 - z}$.
Proof
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 $\size z < 1$.
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:
- $\ds \sum_{n \mathop = 0}^\infty \size z^n = \frac 1 {1 - \size z}$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 6.2$
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.1$: The Geometric Series: $(1.10)$