Compute and simplify the full Taylor series of a function about $x=0$

Question

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? $$f(x)=\frac{1}{2-x}+\frac{1}{2-3x}$$

Answer 1

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.

Answer 2

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.