Define the radius of convergence of the power series by
with when this set is unbounded. It is nonempty because the power series converges at zero.
If , the definition of supremum supplies a point of convergence with . The terms of a convergent series tend to zero and are bounded, say . Therefore
The right-hand side has a convergent geometric series, since . The comparison test proves absolute convergence at .
If and the power series converged at , then would belong to the set whose supremum is , an immediate contradiction. Thus
No assertion about the boundary follows from this argument. If the first assertion is vacuous; if the second is vacuous.

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