Sum of Infinite Geometric Sequence/Proof 3

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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

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

Then:

\(\ds z S\) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty z^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty z^{n + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty z^n\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds S - z^0\)
\(\ds \) \(=\) \(\ds S - 1\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \paren {1 - z} S\)

Hence the result.

$\blacksquare$


Sources

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