Let bound all and put . Then
The geometric series converges, so the comparison test for series proves absolute convergence of . This is the bounded power-series terms force interior absolute convergence principle.
Define the radius of convergence by
allowing and . If , the supremum supplies a point of convergence with ; convergence makes its terms bounded, so the argument above gives absolute convergence at . If , convergence is impossible by definition. At , no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.
Suppose, towards a contradiction, that for every , where . Choose a radius strictly between and , and take any with . Then
The geometric series on the right converges because its ratio lies in . The comparison test for series therefore gives absolute convergence of the power series at , contrary to the defining property of its radius of convergence. Thus is unbounded in whenever . This is the bounded power-series terms force interior absolute convergence principle. The case has no points outside its convergence disk and the assertion is vacuous there.
Every possible radius of convergence is realized by an explicit power series:
For the first, consecutive absolute terms have ratio , so at every they eventually grow and do not tend to zero; at zero the series is defined. The middle geometric series converges absolutely exactly for , and outside that disk its terms grow. The last has consecutive-term ratio , so the ratio test gives absolute convergence for every finite . These arguments use the necessary zero-term condition for convergence, the comparison test for series, and absolute convergence of a geometric series with ratio less than one; they do not assume convergence on the boundary.