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

Theorem

Let $x \in \R$ such that $-1 < x < 1$.

Then:

$\ds \dfrac 1 {\paren {1 + x}^2}$

$=$

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

$\ds $

$=$

$\ds 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - \cdots$

$\ds \frac 1 {1 + x}$

$=$

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

$\ds \leadsto \ \ $

$\ds \frac \d {\d x} \frac 1 {1 + x}$

$=$

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

$\ds \leadsto \ \ $

$\ds -\frac 1 {\paren {1 + x}^2}$

$=$

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

$\ds \leadsto \ \ $

$\ds \frac 1 {\paren {1 + x}^2}$

$=$

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

taking one of the $-1$s out

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} k x^{k - 1}$

the term in $k = 0$ vanishes

$\ds $

$=$

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

$\blacksquare$