Maclaurin Series Expansion

[ChaoticLlama]

Given that the Maclaurin series of \(\displaystyle \L\frac{1}{{1 - x}}\) is \(\displaystyle \L\sum\limits_{n = 1}^\infty {x^n }\), what is the power series expansion for \(\displaystyle \L\frac{{x^2 }}{{1 - x^2 }}\)

 

[skeeter]

actually, the Maclaurin series expansion for $\displaystyle \frac{1}{1-u}$ is
$\displaystyle 1 + u + u^2 + u^3 + ...$

$\displaystyle \frac{x^2}{1 - x^2} =$
$\displaystyle x^2 \frac{1}{1-x^2} =$
$\displaystyle x^2[1 + x^2 + x^4 + x^6 + ...] =$
$\displaystyle x^2 + x^4 + x^6 + ... = \frac{1}{1-x^2} - 1$