Sum of Infinite Geometric Sequence/Corollary 1

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Corollary to Sum of Infinite Geometric Sequence

Let $S$ be a standard number field, i.e. $\Q$, $\R$ 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:

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


Proof 1

\(\ds \sum_{n \mathop = 1}^\infty z^n\) \(=\) \(\ds -z^0 + \sum_{n \mathop = 0}^\infty z^n\)
\(\ds \) \(=\) \(\ds -1 + \frac 1 {1 - z}\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \frac {z - 1 + 1} {1 - z}\)
\(\ds \) \(=\) \(\ds \frac z {1 - z}\)

$\blacksquare$


Proof 2

\(\ds \sum_{n \mathop = 1}^\infty z^n\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty z \cdot z^{n - 1}\)
\(\ds \) \(=\) \(\ds z \sum_{n \mathop = 1}^\infty z^{n - 1}\)
\(\ds \) \(=\) \(\ds z \sum_{m \mathop = 0}^\infty z^m\) setting $m = n - 1$
\(\ds \) \(=\) \(\ds z \frac 1 {1 - z}\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \frac z {1 - z}\)

$\blacksquare$