Compute and simplify the full Taylor series of a function about $x=0$ I was given a function (I will put it in below) and asked to find the Taylor series for it about $x=0$. I'm not sure what I am supposed to do? Should I start finding the derivatives to write it out and then look for a pattern? Is there a different method that I am missing?
I was also told to find the domain of convergence. Can that be found on the domain that the numbers being added keep getting smaller and smaller - so that they will end up as a finite number? $$f(x)=\frac{1}{2-x}+\frac{1}{2-3x}$$
 A: You can do manipulations like $\frac{1}{2-x} = \frac{1/2}{1 - \frac{x}{2}}$ in order to make use of J.W. Tanner's hint.
A: You have that
$$\frac{1}{{2 - x}} = \frac{1}{{2\left( {1 - \frac{x}{2}} \right)}}$$
thus, as long as 
$$\left| {\frac{x}{2}} \right| < 1$$
you can write
$$\frac{1}{{2\left( {1 - \frac{x}{2}} \right)}} = \frac{1}{2}\sum\limits_{n = 0}^\infty  {{{\left( {\frac{x}{2}} \right)}^n}}. $$
Moreover you  have that
$$\frac{1}{{2\left( {1 - \frac{{3x}}{2}} \right)}} = \frac{1}{2}\sum\limits_{n = 0}^\infty  {{{\left( {\frac{{3x}}{2}} \right)}^n}} $$
as long as
$$\left| {\frac{{3x}}{2}} \right| < 1.$$
Therefore as long as
$$\left| x \right| < \frac{2}{3}$$
you have that both the developments hold and  thus you have that
$$\frac{1}{{1 - 3x}} + \frac{1}{{2 - 3x}} = \frac{1}{2}\left( {\sum\limits_{n = 0}^\infty  {{{\left( {\frac{x}{2}} \right)}^n} + \sum\limits_{n = 0}^\infty  {{{\left( {\frac{{3x}}{2}} \right)}^n}} } } \right) = \sum\limits_{n = 0}^\infty  {\frac{{1 + {3^n}}}{{{2^{n + 1}}}}{x^n}}. $$
The last series is the required Taylor series.
