Division of power series Let $f(x) =\sum_{n=0}^{\infty} a_n x^n$ and $g(x) =\sum_{n=0}^{\infty} b_n x^n$ in an open interval $(-r, r)$, and $g \neq 0 $ in $(-r, r)$. How can I prove that $h = f/g$ can be expressed as a power series $\sum_{n=0}^{\infty} c_n x^n$ that converges at least in $(-r,r)$?
I know how to 'calculate' the power series expansion of $h$, for example I can use the determinant method in https://en.wikipedia.org/wiki/Power_series#Multiplication_and_division. The question is, how can I prove that the series expansion 'does exist'(that is, converges)?
 A: This is a long comment and should be treated as such.

Well, it turns out that the radius of convergence of $1/g$ depends not only on that of $g$ but also on the region where $g$ does not vanish.
Consider $$g(x) =2-\cos x=1+\frac{x^2}{2!}-\frac{x^4}{4!}+\dots$$ which converges everywhere so that radius of convergence is $\infty$. But its reciprocal $$\frac{1}{g(x)}=1-\frac{x^2}{2}+\dots$$ does not have radius of convergence $\infty $ but actually it is much smaller.
Why? We know that $g(0)$ is non zero and for real $x$ the value $g(x) >0$, but the reals don't count too much and we exactly need to know a number $R>0$ such that $g(z) \neq 0$ for all complex $z$ with $|z|<R$. Then the radius of convergence of $1/g$ will be at least $R$.
The theorems of Apostol's book also use complex numbers to deal with power series.
I am not very much familiar with complex analysis, but the key is that the idea of a function being expressed via power series in its region of convergence is completely equivalent to a function being analytic in a circular region. I hope there is some textbook available which describes analytic function theory starting from power series approach.
Note: This also explains the term radius of convergence as there is actually a circular region  of convergence in the complex plane involved. When restricting to real variables the only part seen is the interval of convergence and the term radius looks inappropriate for indicating the size of that interval.
A: If there is a purely real analytic approach to finding the radius of convergence of the quotient of two power series, it would be interesting to see.  However, the only approach I know relies on Cauchy's formula, so it requires a bit of complex analysis. The fact that the radius of convergence is equal to the distance from the center of the expansion to the nearest singularity in $\mathbb{C}$, indicates some dependence on complex analysis.
Here is a proof that the power series for an analytic function has a radius of convergence equal to the distance from the center of the expansion to the closest singularity.

Let $R$ be the radius of convergence of
$$
f(z)=\sum_{k=0}^\infty a_n(z-z_0)^n\tag1
$$
where $f$ is analytic inside, and continuous on, $B(r,z_0)=\{z:|z-z_0|\le r\}$. Set $\gamma=z_0+re^{i\theta}$, then
$$
\begin{align}
\frac1R
&=\limsup_{n\to\infty}|a_n|^{1/n}\tag2\\
&=\limsup_{n\to\infty}\left|\frac{f^{(n)}(z_0)}{n!}\right|^{1/n}\tag3\\
&=\limsup_{n\to\infty}\left|\frac1{2\pi i}\oint_\gamma\frac{f(z)}{(z-z_0)^{n+1}}\,\mathrm{d}z\right|^{1/n}\tag4\\[3pt]
&\le\frac1r\tag5
\end{align}
$$
Explanation:
$(2)$: the Cauchy–Hadamard theorem (real analysis)
$(3)$: Taylor's Theorem (real analysis)
$(4)$: Cauchy's Integral Formula (complex analysis)
$(5)$: $|z-z_0|=r$ on $\gamma$, the length of $\gamma$ is $2\pi r$, and $|f|$ is bounded
That is, $R\ge r$, where $f$ is analytic inside, and continuous on, $B(r,z_0)$.

Thus, if the series for $f(z)$ and $g(z)$ converge inside $B(R,z_0)$ and and $g(z)\ne0$ on $B(R,z_0)$, then both $f(z)$ and $1/g(z)$ are analytic inside $B(R,z_0)$, and neither have poles inside $B(R,z_0)$. Therefore, the series for $f/g(z)$ converges inside $B(R,z_0)$.
