What is the radius of convergence of $\frac{1}{1+x^2}$ at $x_0=0$? Which is the radius of convergence of Taylor series of $f(x)=\frac{1}{1+x^2}$ around $x_0=0$? I am unable to write down the analitycal expression of all its derivatives. Why is it  finite even if $f(x)<\infty$ for all $x\in \mathbb{R}$?
 A: One possibility (which does not require complex analysis) is to use the geometric series, as suggested by anon. We know that for $|r|<1$
$$1+r+r^2+\ldots=\sum_{k=0}^\infty r^k=\frac{1}{1-r}.$$
If we use this with $r=-x^2$, we have
$$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\sum_{k=0}^\infty (-x^2)^k=\sum_{k=0}^\infty (-1)^kx^{2k},$$
which is the Taylor series around $x_0=0$ (also known as the Maclaurin series).
From the properties of the geometric series, we know that this series converges iff $|-x^2|<1$. This is equivalent to $|x|<1$, so the radius of convergence is one.
A: The radius of convergence is the distance to the nearest singularity of the function.
Your function has singularities at $x=\pm i$, so the Maclaurin expansion (Taylor series at zero) of $f(x)=\frac{1}{1+x^2}$ has radius of convergence $r=1$ (the distance from zero to $\pm i$).
If the Taylor series expansion is about $x_0=1$, then the radius of convergence would be
$$\min_{x=\pm i}|1-x|=\sqrt{2}.$$
A: $f(\sqrt{-1}) = 1/0$
$f(-\sqrt{-1}) = 1/0$
so the radius of convergence is blocked by these explosions.
