Power Series Expansion for Reciprocal of Square of 1 - z
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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 1
| \(\ds \sum_{n \mathop = 0}^\infty z^n\) | \(=\) | \(\ds \frac 1 {1 - z}\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {\dfrac \d {\d z} } {\sum_{n \mathop = 1}^\infty z^n}\) | \(=\) | \(\ds \dfrac \d {\d z} \frac 1 {1 - z}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {1 - z}^2}\) | Power Rule for Derivatives and the Chain Rule for Derivatives |
Now we have:
| \(\ds \) | \(\) | \(\ds \map {\frac \d {\d z} } {\sum_{n \mathop = 0}^\infty z^n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\frac \d {\d z} z^n}\) | Derivative of Absolutely Convergent Series | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty n z^{n - 1}\) | Power Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty n z^{n - 1}\) | as the zeroth term vanishes when $n = 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n\) | Translation of Index Variable of Summation |
Hence the result.
$\blacksquare$
Proof 2
| \(\ds \dfrac 1 {\paren {1 - z}^2}\) | \(=\) | \(\ds \paren {\dfrac 1 {1 - z} }^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{n \mathop = 0}^\infty z^n}^2\) | 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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n\) |
$\blacksquare$