Question: 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 *n*th 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..

Answer: 
You say that you know $\sum\limits\_{n=0}^{\infty}x^n=\frac{1}{1-x}$. What happens if you multiply both sides by $x^2$?
Think about it for a minute - $-x-1$ is **already** an infinite series $\sum\_{n=1}^\infty a\_nx^n$; in this case, $a\_0 = -1$ and $a\_1 = -1$ and $a\_n=0$ for $n>1$.

(The fact that the series is McLaurin lies in that we are seeking a series of the form $\sum\_{n=1}^\infty a\_nx^n$; for other Taylor expansions we want a series of the form $\sum\_{n=1}^\infty a\_n(x-a)^n$ for some $a$).