Boundary convergence of a power series

Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.