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

Lee David Chung Lin
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Aruha
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  • for $n\gg 1$ the coefficient is dominated by $2(-1)^n$ since $2^{-n}\to 0$ so the radius is the same as $\sum x^n$ which is $1$. – zwim Feb 12 '20 at 18:10
  • What could be the interval? would that be (1, 0.5)? – Aruha Feb 12 '20 at 18:13
  • There are chances it is $(-1,2)$ the points that annulates denominator but it would be better to recenter in $x_0=0.5$ and calculate the new radius of convergence. – zwim Feb 12 '20 at 18:15
  • Great, thanks. This helps. I had confusion around 0.5. Thanks a lot. – Aruha Feb 12 '20 at 18:20

2 Answers2

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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 $(-1,1)$.

Markus Scheuer
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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, $a=0$.

PrincessEev
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