$x\mapsto \frac{1}{1+x^2}$ is analytic Show that $f:\mathbb{R}\to\mathbb{R}, f(x)=\frac{1}{1+x^2}$ is analytic, i.e. that $\forall x_0 \in \mathbb{R}$, $f$ can be approximated in a neighbourhood of $x_0$ by a power series centred at $x_0$.

Is it possible to explicitly find such a power series? I tried some transformations using the geometric series, but without success.
Note that I do not yet know that holomorphic functions are analytic.
This question has been asked before, but I do not understand the solution.
 A: Note that $$ \frac1{1+\alpha x} = \frac1{1+\alpha x_0+\alpha(x-x_0)} =\frac1{1+\alpha x_0}\times \frac1{1+\frac{\alpha(x-x_0)}{1+\alpha x_0}} = \sum_{n\ge 0} \frac{(-1)^n\alpha ^n}{(1+\alpha x_0)^{n+1}}(x-x_0)^n$$
Using this for $\alpha = \pm i$ and the expansion $$\frac1{1+x^2} = \frac{1/2}{1+(-i)x} + \frac{1/2}{1+ix}$$
we have
$$ \frac1{1+x^2} = \sum_{n\ge 0}\left(\frac{i^n}{2(1-ix_0)^{n+1}} + \frac{(-i)^n}{2(1+ix_0)^{n+1}}\right)(x-x_0)^n = \sum_{n\ge 0} \Re\left[\frac{i^n}{(1-ix_0)^{n+1}}\right](x-x_0)^n$$
As a sanity check- if $x_0=0$, then we should see that only the even terms survive. This is in fact true since $i^n/(1-ix_0)^{n+1} = i^n$ is purely real if $n$ is even and purely imaginary if $n$ is odd. Further the coefficient of $x^2$ should be $-1$, and $i^2 = -1$.
I would presume the comment of Jyrki is equivalent.

A small update no one asked for (but I couldn't just leave it at Jyrki's comment...maybe he was thinking about something easier) First note that
$$ \frac{a+ib}i = b -ia. \implies \Re(\frac{z}i) = \Im z. $$
Applying this to $z=i^{n+1}/(1-ix_0)^{n+1}$,
$$\Re\left[\frac{i^n}{(1-ix_0)^{n+1}}\right] = \Im\left[\left(\frac{i}{1-ix_0}\right)^{n+1}\right] = \Im\left[\left(\frac{-1}{i+x_0}\right)^{n+1}\right] = \frac{(-1)^{n+1}}{d(i,x_0)^n}\Im e^{-i n \theta}=\frac{(-1)^{n}\sin(n \theta)}{d(i,x_0)^n} $$
where $d(i,x_0)=\sqrt{1+x_0^2}$ is the distance between $x_0$ and $i$, and $\theta = \arg(i+x_0) = \arctan(1/x_0)$ is the angle that $i+x_0$ makes with the real axis as a vector. 
This isn't really a simpler formula, but its more geometrically meaningful and makes it  mechanically routine to apply the root test for the radius of convergence; there is $n_k\to\infty$ such that $|\sin (n_k\theta)| \in (1/2,1]$, as $\theta$ is never a multiple of $\pi$. Therefore 
$$ \limsup_{n\to\infty} \frac{|\sin(n \theta)|^{1/n}}{d(i,x_0)} \ge \lim_{k\to\infty} \frac{2^{-1/n_k}}{d(i,x_0)} = \frac1{d(i,x_0)}$$
But $\limsup_{n\to\infty} \frac{|\sin(n \theta)|^{1/n}}{d(i,x_0)} \le \frac{1}{d(i,x_0)} $ trivially, so the radius of convergence is exactly $d(i,x_0)$, as standard complex analysis might also give.
