Power Series Expansion for Reciprocal of Square of 1 - z/Proof 2
Jump to navigation
Jump to search
Theorem
| \(\ds \dfrac 1 {\paren {1 - z}^2}\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + 2 z + 3 z^2 + 4 z^3 + \cdots\) |
Proof
| \(\ds \dfrac 1 {\paren {1 - z}^2}\) | \(=\) | \(\ds \paren {\dfrac 1 {1 - z} }^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{n \mathop = 0}^\infty z^n}^2\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\sum_{k \mathop = 0}^n 1 \times 1} z^n\) | Product of Absolutely Convergent Series | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n\) |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.4$. Power Series: Example