<s>[INST] 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.
 [/INST] 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 $\sqrt{a^2+1}$.
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