Power Series Expansion for Reciprocal of 1 + x/Proof 1

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

\(\ds \frac 1 {1 + x}\)

\(=\)

\(\ds \frac 1 {1 - \paren {-x} }\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-x}^k\)

Sum of Infinite Geometric Sequence

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\)

$\blacksquare$