# Proving that a function is real-analytic

I try to solve the following exercise:

Let $f:\mathbb{R}\to\mathbb{R}$ with $f(x):=\frac{1}{1+x^4}$. Prove that $f(x)$ is real analytic and compute the radius of convergence of it's Taylor series at any arbitrary point aroud $x_0\in\mathbb{R}$.

I know that an analytic function is locally equal to a power series. So I thought that the Taylor series is $f(x):=\frac{1}{1+x^4}=\sum\limits_{n=0}^\infty (-x^4)^n$ and it should converges for $|x|<1$, but I don't know if it is correct or how else I should do this exercise, since we have a real function so I can not use Cauchy-Riemann Equation or any other complex theorem.

Thanks for help

Extend your function to $\mathbb C\setminus\{\frac{\pm 1\pm i}{\sqrt2}\}\to \mathbb C$ by defining $f(z) = \frac1{1+z^4}$.