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

Corollary to Power Series Expansion for $\frac 1 {\paren {1 + x} }$

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

Then:

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

$=$

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

$\ds $

$=$

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

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \paren {-1}^{2 k} \paren {k + 1} x^k$

$\ds $

$=$

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

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Negative and fractional indices