Proof. The radius of convergence is \(1\) because the coefficients are \(0\) and \(1\) and infinitely many are \(1\). The series satisfies the functional equation \[f(z) = z + f(z^2),\] and iterating it \(m\) times gives \[f(z) = z + z^2 + \cdots + z^{2^{m-1}} + f\big (z^{2^m}\big ).\] Let \(\zeta \) be any \(2^m\)-th root of unity, so that \(\zeta ^{2^m} = 1\). Along the radius \(z = r\zeta \) with \(r \rightarrow 1^{-}\) the last term is \(f(r^{2^m})\), which tends to \(+\infty \) because \(f\) has non-negative coefficients and \(f(r) \rightarrow \infty \) as \(r \rightarrow 1^{-}\). The remaining terms stay bounded, so \(\left |f(r\zeta )\right | \rightarrow \infty \) and \(\zeta \) is a singular point.

The \(2^m\)-th roots of unity, taken over all \(m\), are dense in the unit circle. Every arc of the circle therefore contains a singular point, and no arc admits a continuation. \(\blacksquare \)

So the circle of convergence is sometimes an artefact of the formula — as for the geometric series — and sometimes a genuine wall. Nothing about the radius alone distinguishes the two cases. □