Sum of Infinite Geometric Sequence/Corollary 1/Proof 2

Corollary to Sum of Infinite Geometric Sequence

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

Proof

\(\ds \sum_{n \mathop = 1}^\infty z^n\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty z \cdot z^{n - 1}\)

\(\ds \)

\(=\)

\(\ds z \sum_{n \mathop = 1}^\infty z^{n - 1}\)

\(\ds \)

\(=\)

\(\ds z \sum_{m \mathop = 0}^\infty z^m\)

setting $m = n - 1$

\(\ds \)

\(=\)

\(\ds z \frac 1 {1 - z}\)

Sum of Infinite Geometric Sequence

\(\ds \)

\(=\)

\(\ds \frac z {1 - z}\)

$\blacksquare$