<s>[INST] Finding the radius of convergence for $\frac{1}{(3 - x)(1 + x)^{2}}$ at $x_{0} = 0$ and $x_{0} = 1$ I am having trouble with the problem 

Find the radius of convergence for $f(x) = \frac{1}{(3 - x)(1 +
 x)^{2}}$ at $x_{0} = 0$ and $x_{1} = 1$.

Usually, when I'm trying to find radius of convergence, I'm given some sort of series (like, a summation). This isn't the case here, and I'm confused as to what I'm supposed to do. This is a problem in the exercise. It is in the section for power series. Also, I have this formula for the radius of convergence $R$:
$$R = \frac{1}{\limsup_{n\to\infty} |a_{n}|^{1/n}},$$
but I'm not sure how I'm supposed to use that here either. 
Any help is appreciated.
EDIT: Maybe it has something to do with the standard Taylor series form? Like, 
$$f(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + f''(x_{0})(x - x_{0})^{2}/2! \ldots $$
Also, in case it matters, by Partial Fractions, you can write
$$\frac{1}{(3-x)(1 + x)^{2}} = \frac{1}{16(x + 1)} + \frac{1}{4(x+1)^{2}} - \frac{1}{16(x - 3)}. $$
The first few terms of the Taylor series expansion about $x_{0} = 0$ is given by
$$f(x) = \frac{1}{3} - \frac{5}{9}x + \frac{22}{27}x^{2} - \frac{86}{81}x^{3} + \frac{319}{243}x^{4} + \cdots $$
(I don't have a closed term for the sum, but these are the first terms explicitly computed. Looks like the denominator is just powers of $3$. Not sure about the numerator though...)
 [/INST] Here are my thoughts. This might not be the original purpose of the exercise, but this is my guess.
According to your partial fraction decomposition, and using the well-know power series $(1-x)^{-1} = \sum_0^\infty x^n$ for $x \in (-1,1)$, at $x=0$, we have
\begin{align*}
f(x)&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac {\mathrm d}{\mathrm dx} \left( \frac 1{4(x+1)} \right) + \frac 1{48} \cdot \frac 1{1 - x/3} \\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14 \cdot\frac {\mathrm d }{\mathrm dx}\left( \sum_0^\infty (-1)^n x^n \right) + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14  \sum_0^\infty (n+1)(-1)^{n+1} x^{n}  + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \sum_0^\infty \left(\frac {(-1)^n}{16} + \frac 14 (n+1)(-1)^{n+1} +\frac 1{16 \cdot 3^{n+1}} \right) x^n,
\end{align*}
and all of these computations are valid iff $|x|<1$ and $|x/3|<1$. Conclusively the radius  is $1$. You might try to compute again by the formula. 
At $x=1$, just let $y = x-1$, then $x = y+1$ and 
$$
f(x) =g(y) = \frac 1{(2-y)(2+y)^2},
$$ 
you could do the similar thing like above to give the power series at $y=0$. You could see that the operations are valid iff $|y/2|<1$, hence $|y|<2$ and the radius is $2$. 
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