Radius of convergence via the ratio test

Fix $x \neq c$ and apply the ratio test to the terms $b_n = a_n(x-c)^n$ of the numerical series $\sum b_n$. We compute

$\displaystyle \lim_{n\to\infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n} \right| = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \cdot |x-c| = L\,|x-c|$.

By the ratio test for numerical series, $\sum b_n$ converges absolutely when $L|x-c| < 1$, i.e. when $|x-c| < 1/L$, and diverges when $L|x-c| > 1$, i.e. when $|x-c| > 1/L$. So the series converges absolutely for every $x$ with $|x-c| < 1/L$ and diverges for every $x$ with $|x-c| > 1/L$.

This is exactly the definition of the radius of convergence: the largest $R$ such that the series converges for all $|x-c| < R$. Hence $R = 1/L$. The boundary cases $L=0$ (ratio always shrinks to 0, so the series converges for every $x$, giving $R=\infty$) and $L=\infty$ (ratio blows up for any $x \neq c$, so the series converges only at $x=c$, giving $R=0$) follow the same argument taking the appropriate limits.