Power Series Expansion for Reciprocal of 1 + x/Proof 2
Jump to navigation
Jump to search
Theorem
Let $x \in \R$ such that $-1 < x < 1$.
Then:
\(\ds \dfrac 1 {1 + x}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - x + x^2 - x^3 + x^4 - \cdots\) |
Proof
\(\ds \frac 1 {1 + x}\) | \(=\) | \(\ds \paren {1 + x}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^{\underline k} } {k!} x^k\) | General Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-1} - j} } {k!} x^k\) | Definition of Falling Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 1} } } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Definition of Factorial and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-j} } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Translation of Index Variable of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k j} {\ds \prod_{j \mathop = 1}^k j} x^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\) | simplification |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Negative and fractional indices
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Binomial Series: $20.8$