Radius of convergence for $\frac{1}{x^2 + x - 6}$

Question

The problem I'm working on asks for the radius of convergence for the Maclaurin series of $\frac{1}{x^2 + x - 6}$

Here is my attempt at a solution. I would highly appreciate anyone who spots the error(s).

By partial fraction decomposition we have

$$\frac{1}{x^2 + x - 6} = \frac{\frac{1}{5}}{x - 2} - \frac{\frac{1}{5}}{x + 3}$$

$$ = \biggl(\frac{1}{5}\biggl)\frac{1}{(-2)(1 -\frac{x}{2})} - \biggl(\frac{1}{5}\biggl)\frac{1}{(-3)(1 -\frac{x}{3})} $$

$$ = \biggl(\frac{1}{5}\biggl) \biggl[ \biggl(-\frac{1}{2}\biggl) \frac{1}{(1 -\frac{x}{2})} + \biggl(\frac{1}{3}\biggl)\frac{1}{(1 -\frac{x}{3})} \biggl]$$

Converting to power series and absorbing the fractions into the respective summations:

$$ = \biggl(\frac{1}{5}\biggl)\biggl( \sum_{n=0} ^{\infty} \frac{x^n}{3^{n + 1}} -\sum_{n=0} ^{\infty} \frac{x^n}{2^{n + 1}} \biggl)$$

Combining summations and computing a common denominator we have

$$ = \biggl(\frac{1}{5}\biggl)\biggl( \sum_{n=0} ^{\infty} \frac{x^n(2^{n + 1} - 3^{n + 1})}{3^{n + 1}2^{n + 1}} \biggl)$$

We now apply the ratio test to the general term of our power series:

$$ \lim_{n\to\infty} \biggl|\frac{x^{n + 1}(2^{n + 2} - 3^{n + 2})}{3^{n + 2}2^{n + 2}} \cdot \frac{3^{n + 1}2^{n + 1}}{x^n(2^{n + 1} - 3^{n + 1})} \biggl| $$

$$ = \lim_{n\to\infty} \biggl|\frac{x^{n + 1}(2^{n + 2} - 3^{n + 2})}{3^{n + 2}2^{n + 2}} \cdot \frac{3^{n + 1}2^{n + 1}}{x^n(2^{n + 1} - 3^{n + 1})} \biggl| $$

$$ = \frac{x}{6} \biggl( \lim_{n\to\infty} \biggl|\frac{2^{n + 2}}{2^{n + 1} - 3^{n + 1}} \biggl| - \lim_{n\to\infty} \biggl|\frac{3^{n + 2}}{2^{n + 1} - 3^{n + 1}}\biggl| \biggl)$$

Throwing the $\frac{1}{5}$ back in we get:

$$ \biggl(\frac{1}{5} \biggl) \biggl(\frac{x}{6}\biggl) (0 - 3)$$

$$ \Longrightarrow $$

$$ \biggl|- \frac{x}{90}\biggl| < 1 $$

So $ x < 90 $.

But the radius of convergence is supposed to be $ 2 $ so this is clearly wrong.

(As a humorous aside, ChatGPT was unable to go through these steps without making numerous errors, often arriving at a completely different answer when asked if it might have made a mistake).

Answer 1

Combining summations and computing a common denominator we have

$$ \biggl( \sum_{n=0} ^{\infty} \frac{1}{5}\frac{x^n(2^{n + 1} - 3^{n + 1})}{3^{n + 1}2^{n + 1}} \biggl)$$

We now apply the ratio test to the general term of our power series:

$\begin{align} \lim_{n\to\infty} \biggl|\frac{1}{5}\frac{x^{n + 1}(2^{n + 2} - 3^{n + 2})}{3^{n + 2}2^{n + 2}} \cdot \frac{5}{1}\frac{3^{n + 1}2^{n + 1}}{x^n(2^{n + 1} - 3^{n + 1})} \biggl|&=\lim_{n\to\infty} \biggl|\frac{x(2^{n + 2} - 3^{n + 2})}{6(2^{n + 1} - 3^{n + 1})} \biggl|\\ &=\lim_{n\to\infty} \biggl|\frac{x(\frac{2^{n + 2}}{3^{n + 2}}-1)3^{n+2}}{6(\frac{2^{n + 2}}{3^{n + 2}}-1)3^{n+1}} \biggl| \end{align}$

and we get $$ \frac{|x|}{2}\lim_{n\to\infty} \biggl|\frac{\frac{2^{n + 2}}{3^{n + 2}}-1}{\frac{2^{n + 1}}{3^{n + 1}}-1} \biggl|=\frac{|x|}{2}<1. $$ Also, you didn't need to combine the series: The radius of convergences of the series $\sum_{n=0} ^{\infty} \frac{x^n}{2^{n + 1}}$, $\sum_{n=0} ^{\infty} \frac{x^n}{3^{n + 1}}$ are $2$ and $3$ respectively and we have to take the minumum to find the radius of the convergence of their difference.

Answer 2

Indeed, since $\displaystyle \sum_{0}^{+\infty}x^k=\frac{1}{1-x}$ with convergence for $|x|<1$ so one can write

\begin{align*} \frac{1}{x^2+x-6}&=\frac{1/5}{x-2}+\frac{-1/5}{x+3}\\ &=\frac{1}{5}\sum_{0}^{+\infty}(-\frac{1}{2})(\frac{x}{2})^k-\frac{1}{5}\sum_{0}^{+\infty}\frac{(-1)^k}{3} (\frac{x}{3})^k\\ \end{align*}

The first series converges for $|x|<2$ and the second series converges to $|x|<3$ so the sums of the series should be converge for $|x|<2$ (because between $2$ and $3$ diverges). One can see it as follows

Adding series over its intervals of convergence one can write

\begin{align*} \frac{1}{5}\sum_{0}^{+\infty}(\frac{-1}{2^{k+1}}+\frac{(-1)^{k+1}}{3^{k+1}})x^k&=\frac{1}{5}\sum_{0}^{+\infty}(-(\frac{1}{2})^{k+1}+(-\frac{1}{3})^{k+1})x^k\\ &:=\frac{1}{5}\sum_{0}^{+\infty}c_kx^k \end{align*}

Since one has that
\begin{align*} |\frac{c_{k}}{c_{k+1}}|&=|\frac{-(\frac{1}{2})^{k+1}+(-\frac{1}{3})^{k+1}}{-(\frac{1}{2})^{k+2}+(-\frac{1}{3})^{k+2}}|\\ &\underset{k\to+\infty}{\longrightarrow \frac{1}{1/2}}=2 \end{align*}

Since the limit there exists and it is $2$ so the radius of convergence is the limit, that is, $2$.

NB: The Maclaurin series is Taylor series centered at $a=0$. The map $x\mapsto \frac{1}{x^2+x-6}$ has singularity when $x^2+x-6=0$ so $x_1=-3$ and $x_2=2$. Since

The radius of convergence of a power series $f$ centered on a point $a$ is equal to the distance from $a$ to the nearest point where $f$ cannot be defined in a way that makes it holomorphic. (cf.https://en.m.wikipedia.org/wiki/Radius_of_convergence)

Hence the radius of convergence is $R=2$, again.