Taylor Series Expansion of $\frac{1}{1+x^2}$ about $x=a$ Let $$f(x)=\frac{1}{1+x^2}$$ Consider its Taylor series expansion about a point $a\in \mathbb{R}$. What is the radius of convergence of this series??
About $x=0$ we could expand it like 
$$(1+x^2)^{-1}= 1-x^2+(-1)(-2) \frac{x^2}{2!}+(-1)(-2)(-3)\frac{x^3}{3!}+\dots$$ and get the radius as $1$.
But about $x=a$, there is a slight problem with the way as above...
$$\frac{1}{1+(x-a+a)^2}=\frac{1}{1+a^2(\frac{x-a}{a}+1)^2}.$$
Here is where I am getting stuck.

Here is a pretty simple way to get the actual Taylor series centered at $a$:
Using partial fractions:
$\dfrac{1}{z^2+1} = \dfrac{i}{2}\left(\dfrac{1}{z+i} - \dfrac{1}{z-i}\right)$.
Now, we can expand each of the the two using the geometric series:
$\displaystyle\dfrac{1}{z+i} = \dfrac{1}{(z-a)+(a+i)} =\dfrac1{a+i}\sum_{n=0}^\infty(-1)^n \left(\dfrac{z-a}{a+i}\right)^{n}$, $|z-a| < |a+i| = \sqrt{1+a^2}$
$\displaystyle\dfrac{1}{z-i} = \dfrac{1}{(z-a)+(a-i)} =\dfrac1{a-i}\sum_{n=0}^\infty(-1)^n \left(\dfrac{z-a}{a-i}\right)^{n} $, $|z-a| < |a-i| = \sqrt{1+a^2}$
$\displaystyle\dfrac{1}{z^2+1} = \dfrac{i}{2}\sum_{n=0}^\infty(-1)^n \left(\left(\dfrac{1}{a+i}\right)^{n+1}-\left(\dfrac{1}{a-i}\right)^{n+1}\right)(z-a)^n$
Which maybe rewritten as:
$\displaystyle\dfrac{1}{z^2+1} = \sum_{n=0}^\infty(-1)^{n+1} Im\left(\left(\dfrac{1}{a+i}\right)^{n+1}\right)(z-a)^n $

EDIT: Radius of convergence from the series.

We can see that the the radius of convergence must be at least $\sqrt{1+a^2}$, since the two series we manipulated had this radius of convergence. We can also deduce this from the following inequality.
$\left|Im\left(\left(\dfrac{1}{a+i}\right)^{n+1}\right)\right| \le \left|\left(\dfrac{1}{a+i}\right)^{n+1}\right|= \left(\dfrac1{\sqrt{1+a^2}}\right)^{n+1}$.
On the other hand, the radius of convergence cannot be lager than $\sqrt{1+a^2}$ because of the singularity of $1/(1+z^2)$.
