Taylor Series and Differentiation with Sigma notation $f(x) = \frac{x}{(2-3x)^2}$ Use Term By Term Differentiation to Find the Taylor Series about $x$=3 for 
Give The Open Interval of Convergence and express as sigma notation 
 $\sum A_n(x-3)^n$
$f(x) = \frac{x}{(2-3x)^2}$
So I have Found the Taylor Series for 1/(2-3x) to be 
$\sum{(-3)^{n}\cdot(x-3)^{n} }\cdot{(-7)^ {n+1}}$
How Do you find the original function taylor series and its interval of convergence and then express in sigma notation of the form  $\sum A_n(x-3)^n$
 A: Since you are concerned by the Taylor series of $$f(x) = \frac{x}{(2-3x)^2}$$ around $x=3$, just as able answered, you can start using $x-3=y$, $x=y+3$, $2-3x=-3y-7$ and consider the  Taylor series of $$f(y) = \frac{y+3}{(3y+7)^2}$$ around $y=0$.
May be, a convenient way could then be to write $$y+3=(3y+7)^2\sum_{i=0}^\infty a_i\,y^i$$ and identify the coefficients. Developing the last expression, we then have $$3+y=9\sum_{i=0}^\infty a_i\,y^{i+2}+42\sum_{i=0}^\infty a_i\,y^{i+1}+49\sum_{i=0}^\infty a_i\,y^i$$ and we need to identify the $a_i$ coefficients.
For the terms of degree $0$ and $1$ present in the lhs, we then have $$3=49a_0\implies a_0=\frac 3{49}$$ $$1=42a_0+49a_1\implies a_1=-\frac{11}{343}$$ For all other terms $$49a_n+42a_{n-1}+9a_{n-2}=0$$ So, we have the recursion for which you must notice that the characteristic equation has a double root equal to $-\frac 37$.
Using the initial conditions, the standard method then leads to $$a_n=\frac{1}{147} \left(-\frac 37\right)^n(2n+9)$$ and finally $$f(x) = \frac{x}{(2-3x)^2}=\frac{1}{147}\sum_{i=0}^\infty \left(-\frac 37\right)^n(2n+9)(x-3)^i$$
A: you can make a change of variable $x = 3 + h.$  then we have 
$$\begin{align} f(x) &= \frac{x}{(2-3x)^2} = \frac{3+h}{(2 - 9-3h)^2}=\frac 1{49}(3+h)\left(1 + \frac{3h}7\right)^{-2} \\
&=\frac 1{49}(3+h)\left(1 -\frac 21 \frac{3h}7 + \frac{2\cdot3}{1\cdot2}\left(\frac{3h}{7}\right)^2 - \frac{2\cdot3 \cdot 4}{1\cdot2\cdot 3}\left(\frac{3h}{7}\right)^2 + \ldots\right)   \end{align} $$
i will let you multiply out the two factors and collect the like terms.
