First off all, I am sorry if my english is not perfect. I need help again for this exercise:
Find Maclaurin series expansion for $f(x)=\frac{x^2}{1-x}$.
That's what I did:
$f(x)=\frac{x^2}{1-x}=\frac{x^2+1-1}{1-x}=\frac{(x+1)(x-1)}{1-x}+\frac{1}{1-x}= -x-1+\frac{1}{1-x}$.
I know that $\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1 $. But What to do with $-x-1$?
(-x-1)'=-1 and (-x-1)''=0.
If I chose to find the nth derivative for $f(x)$:
f'(x)=-1+(1-x)^{-2}
f''(x)=0+(-2)(1-x)^{-3}(-1)=(-1)^22!(1-x)^{-3}
$...................................................$
$f^{(n)}(x)=(-1)^{2n}(1-x)^{-(n+1)}n! \Rightarrow f^{(n)}(0)=n! \Rightarrow f(x)=\sum\limits_{n=0}^{\infty} x^n$
It's this correct? I am a little confused and I hope someone could help me again..