How to attempt power series questions ? I'm confused with this topic 
Use the power series representation of $\frac{1}{1-x}$ to obtain the power series representation of $\frac{2x}{4-3x}$ and the corresponding values of $x$.

Firstly, I tried to make $\frac{2x}{4-3x}$ in the form of $\frac{1}{1-x}$
$\frac{x}{2} \frac{1}{(1-3x/4)} = \frac{x}{2} \sum_{n=0}^{\infty} (\frac{3x}{4})^n $ for $-1 < 3x/4 < 1$
I know that the power series representation of $x=a$ is $\sum_{n=0}^{\infty} C_n (x-a)^n $ where a is the centre. Am I supposed to determine what is $a$?
What am I supposed to do next? I do not understand the question and what formula am I supposed to be following. I just learnt about power series.
The answer is $\sum_{n=0}^{\infty} (\frac{1}{2}) (\frac{3}{4})^n x^{n+1}$ for $-4/3 < x < 4/3$
What is being done above?

Maybe it is not necessary to note that:
Another way is Maclaurin series. $f(x)=\frac{2x}{4-3x}=-\frac{2}{3}+\frac{8}{3}\frac{1}{4-3x}$. With this form we can easily compute derivatives: $n$-th derivative is $f^{(n)}(x)=\frac{8}{3}3^{n}n!(4-3x)^{-n-1}$ and $f^{(n)}(0)=\frac{1}{2}\frac{3^{n-1}}{4^{n-1}}n!$ for $n>0$.
Noting that $f(0)=0$, $f(x)=\sum_{n=1}^{\infty}\frac{1}{2}\frac{3^{n-1}}{4^{n-1}}{n!}\frac{x^n}{n!}=\sum_{n=0}^{\infty}\frac{1}{2}\frac{3^{n}}{4^{n}}x^{n+1}=\frac{x}{2}\sum_{n=0}^{\infty}\frac{3^{n}}{4^{n}}x^{n}.$
For the interval of convergence, we apply the ratio test which gives $|\frac{3}{4}x|<1$ and hence $-\frac{4}{3}<x<\frac{4}{3}.$
Sometimes at boundary points, here they are $x=\pm\frac{4}{3}$, the series may be convergent. But it is not the case for this series.
