Find the radius of convergence of the Taylor series of $\frac {1}{2+x^2}$ about x=2

Question

My attempt: $$\frac {1}{2+x^2}=\frac {1}{6+[(x-2)^2+4(x-2)]}=\frac{1}{6[1-\frac {(-(x-2)^2)-4(x-2)]}{6}]}$$ This final expression can be treated in the form $\frac {1}{6(1-y)}$ where $y=\frac {-(x-2)^2-4(x-2)}{6}$. We know that $$\frac{1}{1-y}=\sum_{n=0}^\infty y^n for |y|<1$$ Thus the required Taylor series should be $$\frac{1}{6}\sum_{n=0}^\infty(\frac {-(x-2)^2-4(x-2)}{6})^n$$ Using binomial expansion this series can be expressed as $$\frac{1}{6}-\frac{1}{9}(x-2)+\frac{5}{108}(x-2)^2+...$$ We know that this series should converge for $|y|<1$ or $$|\frac {-(x-2)^2-4(x-2)}{6}|<1$$ This gives the interval of convergence to be $(-\sqrt10,\sqrt10)$. However if this series was a power series, there should exist a no non-negative real number R such that the series converges for $|x-2|<R$. However I am not able to find any such radius of convergence R for this series. So can I conclude that the series that I have come up with is not a power series and hence not a Taylor series as every Tailor series is a power series?

Answer 1

To compute the series do the partial fraction decomposition and make appear the $x-2$ term: $$\frac{1}{2+x^2}=\frac{-\sqrt{2}i/2}{x-\sqrt{2}i}+\frac{\sqrt{2}i/2}{x+\sqrt{2}i}=\frac{-\sqrt{2}i/2}{(x-2)+2-\sqrt{2}i}+\frac{\sqrt{2}i/2}{(x-2)+2+\sqrt{2}i}$$

Then this is equal to (by taking out the factors) $$\frac{-\sqrt{2}i/2}{2-\sqrt{2}i}\frac{1}{1+(x-2)/(2-\sqrt{2}i)}+ \frac{\sqrt{2}i/2}{2+\sqrt{2}i}\frac{\sqrt{2}i/2}{1+(x-2)/(2+\sqrt{2}i)}$$

Now, use with each fraction that $\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+...$ For the first fraction put $y=(x-2)/(2-\sqrt{2}i)$ and for the second put $y=(x-2)/(2+\sqrt{2}i)$.

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However, you don't need to compute the series to tell the radius of convergence. The series is known to converge from the center to the nearest singularity. This function has singularities at $x=\pm\sqrt{2}i$. The distance from them to $2$ is $|2\pm\sqrt{2}i|=\sqrt{2^2+2}=\sqrt{6}$.