Power Series Expansion of Reciprocal of 1 + x

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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

$\blacksquare$