Interval of convergence of a series How to get the interval of convergence for the given function,
$$f(x) = \frac{1}{2+x-x^2}$$
I have computed the Maclaurin series and the generalized power series as follows however I am unable to proceed with the valid interval of convergence. Kindly help.
$$\frac{1}{2+x-x^2} = \sum_{n=0}^{\alpha }\frac{1}{6}x^n(2(-1)^n+2^{-n}) \\ 
\frac{1}{2+x-x^2}=\frac{1}{2}-\frac{1}{4}x+\frac{3}{4}\frac{x^{2}}{2!}+ \ldots$$
 A: We can also  calculate  the   partial  fraction decomposition 
\begin{align*}
\frac{1}{2+x-x^2}=\frac{1}{3(x+1)}-\frac{1}{3(x-2)}\tag{1}
\end{align*}
and obtain this  way geometric series. We  see the radius  of convergence of  the geometric series  at $x=0$  is $1$ resp. $2$. So,  the  radius of convergence  of the left-hand  side of (1) is the minimum distance from $x=0$ to the  nearest singularity at  $x=-1$  and the interval of convergence is   $(-1,1)$.
A: There are several different ways to find the radius of convergence $r$. Let $f(x) = \sum c_n(x-a)^n$. 
$$r = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| = \lim_{n \to \infty} \frac{1}{\sqrt[n]{|c_n|}}$$
For you, $c_n = (2(-1)^n + 2^{-n})/6$. I would suggest using the first test as a result. As $n$ grows without bound, the $2^{-n(+1)}$ terms will go to zero, and the absolute value will get rid of the $(-1)^{n(+1)}$ terms, letting you reach a conclusion easily.
As for the interval itself, for the general $f$ described above, it is $(a-r,a+r)$. For you, $a=0$.
