TITLE: Is the reciprocal of an analytic function analytic?
QUESTION [12 upvotes]: Let $f(x)=\sum_{n=0}^{\infty}a_nx^n$ be a function on the real line with a positive(or infinite) radius of convergence about $x=0$, and $a_0\neq0$. Then is it true that the function $1/f(x)$ can also be Taylor expanded about the origin? The question is essentially whether the formal inverse series has a positive or infinite radius of convergence.

REPLY [12 votes]: Yes. I think the following is a way to do this without complex analysis (although I think it would be the easiest way).
For simplicity, let's assume $a_0 = 1$ (otherwise you just divide by $a_0$). And take a radius $\rho > 0$ such that $\sum_{n=1}^{\infty} |a_n| |x|^n < 1$ for all $|x| < \rho$ (there exists one since $x \mapsto \sum_{n=1}^{\infty} |a_n| |x|^n$ is continuous, and zero at the origin). Let's denote $b_n$ for the coefficient of the formal inverse and show by recurrence that $|b_n| \le \rho^{-n}$ (which will prove it converges within  the radius $\rho$).
For $n = 0$ it is obvious since $b_0 = 1$. Then for $n \ge 1$, you have
$$b_n = - \sum_{k=1}^{n} a_k b_{n-k}$$
And
$$|b_n| \le \sum_{k=1}^{n} |a_k| \rho^{k-n}$$
So
$$|b_n| \rho^n \le \sum_{k=1}^{n} |a_k| \rho^{k} \le \sum_{k=1}^{\infty} |a_k| \rho^{k} < 1$$
Which concludes.