Interval of convergence of a series

Question

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. $$\frac{1}{2+x-x^2} = \sum_{n=0}^{\alpha }\frac{1}{6}x^n(2(-1)^n+2^{-n}) \\ \frac{1}{2+x-x^2}=\frac{1}{2}-\frac{1}{4}x+\frac{3}{4}\frac{x^{2}}{2!}+ \ldots$$

Answer 1

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, $a=0$.

Answer 2

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 $(-1,1)$.