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.
 A: Since $f$ is analytic in the domain $\mathbb{C} \setminus \{-i,i\}$, its power series around a point $a$ has radius of convergence at least the distance to the boundary of the domain. That is, the nearest of the two points $\{-i,i\}$. If $a$ is real, these distances are equal. Moreover, the radius of convergence cannot be more than this, since $f$ has singularities at $i$ and $-i$. The distance from $a$ to $i$ is $\sqrt{1 + a^2}$ by Pythagorean theorem.
A: With $u=x-a$:
$$
\begin{align}
\frac1{1+(u+a)^2}
&=\frac1{1+a^2+2au+u^2}\\
&=\frac1{1+a^2}\frac1{1+\frac{2au+u^2}{1+a^2}}\\
&=\frac1{1+a^2}\left(1-\frac{2au+u^2}{1+a^2}+\left(\frac{2au+u^2}{1+a^2}\right)^2-\left(\frac{2au+u^2}{1+a^2}\right)^3+\dots\right)\\[4pt]
&=\frac1{1+a^2}-\frac{2a}{(1+a^2)^2}u+\frac{3a^2-1}{(1+a^2)^3}u^2-\frac{4a^3-4a}{(1+a^2)^4}u^3+O\left(u^4\right)
\end{align}
$$
The radius of convergence is not simple to find from this power series, but as J.J. says, the radius of convergence of the power series for an analytic function is the distance from the point of expansion to the nearest singularity.  If $a\in\mathbb{R}$, this is $\sqrt{a^2+1}$.
A: 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 $1/(1+z^2)$.
A: Idea:
$$1+z+z^2+z^3+\cdots={1\over 1-z}=\frac{1}{1+(x-a+a)^2}=\cdots$$
and the radius of convergence of $1+z+z^2+z^3+\cdots$ is (obviously?) 1.
General principle: the radius of convergence will be the distance form $a$ to the nearest singularity in $\Bbb C$ of the function.
