$\frac{1}{1+x^2}$'s Taylor expansion about a point $a\in\mathbb{R} $, given by $f(x) = \sum_{n = 0} ^ \infty$ $a_n (x -a)^n$. Radius of convergence? Let $f(x) = \frac{1}{1+x^2}$. Consider its Taylor expansion about a point $a \in \mathbb{R}$ given by
$$f(x)=\sum_{n = 0}^\infty a_n(x-a)^n.$$
What is the radius of convergence?
My try:
$\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5}..........$
Differentiating both sides I get 
$\frac{1}{1+x^2} = 1 - x^2 +x^4 -.............$
How can I proceed from here?
Can anyone please help me. 
Answer is $\sqrt{a^2+1}$.

I do not know if there is a simple proof by real analytic methods but it is elementary when you consider the complex function $\frac 1 {1+z^{2}}$. The largest disc around a on which the function is analytic has radius $(1+a^{2})^{1/2}$, the distance from a to the points $i,-i$.
