Develop in series Mac-Laurin the function $f(x)={\frac{x^2}{1-x}}$

Question

I just want to know if what I did was good because I did not fully understood this theorem. I have this function: $$f(x)={\frac{x^2}{1-x}}$$ And I have to develop her in series Mac Laurin. Firstly I've used the geometric series for ${\frac{1}{x^n}}$ to do this $f(x)=x^2\sum_{n\ge0}{\frac{(-1)^n}{x^n}}=\sum_{n\ge0}{\frac{(-1)^nx^2}{x^n}}=\sum_{n\ge0}{\frac{(-1)^n}{x^{n-2}}}=\sum_{n\ge0}{\frac{1}{(-x)^{n-2}}}$ It is correct what I did? If yes, is this the final form of the series Mac-Laurin or shoul I do something more to it?

Answer 1

The MacLaurin series representation for $\frac1{1-x}$ is given by

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

for $|x|<1$. The is simply a geometric series.

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Hence, the series representation for $f(x)=\frac{x^2}{1-x}$ is

$$\begin{align} f(x)&=\frac{x^2}{1-x}\\\\ &=x^2\frac{1}{1-x}\\\\ &=x^2\sum_{n=0}^\infty x^n\\\\ &=\sum_{n=0}^\infty x^{n+2}\\\\ &=\sum_{n=2}^\infty x^{n} \end{align}$$

for $|x|<1$.

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For $|x|>1$, we can write

$$\begin{align} f(x)&=\frac{x^2}{1-x}\\\\ &=-\frac{x}{1-1/x}\\\\ &=-x\sum_{n=0}^\infty\left( \frac{1}{x}\right)^n\\\\ &=-\sum_{n=0}^\infty\left( \frac{1}{x}\right)^{n-1}\\\\ &=-\sum_{n=-1}^\infty x^{-n} \end{align}$$