# Power Series Expansion of Reciprocal of 1 + x

## 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 1

 $\ds \frac 1 {1 + x}$ $=$ $\ds \frac 1 {1 - \paren {-x} }$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \paren {-x}^k$ Sum of Infinite Geometric Sequence $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k$

$\blacksquare$

## Proof 2

 $\ds \frac 1 {1 + x}$ $=$ $\ds \paren {1 + x}^{-1}$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^{\underline k} } {k!} x^k$ 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

$\blacksquare$