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The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

For a series ${\sum }_n{a}_n{x}^{m_n}$ with distinct increasing integer exponents, define the ordinary coefficients to be $a_n$ at $m_n$ and zero elsewhere. The Cauchy-Hadamard theorem then givesThe exponent in the root is the actual degree $m_n$, not the index $n$. For $a_n=(n!{)}^2$ and $m_n=n^2$, $0\le 2\log (n!)/n^2\le 2\log n/n\to 0$, so $R=1$. Boundary behavior is a separate question.

On ${\mathbb{C}}_p$, the power series $\exp (x)={\sum }_{n\ge 0}{x}^n/n!$ has radius of convergenceIndeed, Legendre formula gives $v_p(n!)/n\to 1/(p-1)$, and the Cauchy-Hadamard theorem gives the stated radius.

The generating function weights each self-avoiding walk by $x$ to the power of its length, including the zero-length walk. Its radius of convergence is the reciprocal of the upper exponential growth rate of the rooted counts, by the Cauchy-Hadamard theorem.