= Boundary convergence of a power series
Knowing the <radius of convergence> does not settle convergence on its boundary. The three <power series>
$$
\sum_{n=0}^{\infty}z^n,\qquad
\sum_{n=1}^{\infty}\frac{z^n}{n},\qquad
\sum_{n=1}^{\infty}\frac{z^n}{n^2}
$$
all have radius one, but behave differently on $|z|=1$.
\Image[/power-series-boundary-behavior.png]
{title=The same <radius of convergence> permits three different behaviours on the <unit circle>}
{description=Green denotes <absolute convergence>, orange <conditional convergence>, and red divergence. All three series converge absolutely in the open <unit disk>. On its boundary, the geometric series diverges everywhere; the $1/n$ series converges conditionally except at $z=1$; the $1/n^2$ series converges absolutely everywhere.}
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> obey
$$
\left|\sum_{k=1}^{N}z^k\right|
=\left|\frac{z(1-z^N)}{1-z}\right|
\le\frac2{|1-z|}.
$$
The <Dirichlet test> with the decreasing <sequence> $1/n\to0$ therefore gives <conditional convergence>. In particular, $z=-1$ gives the negative of the <alternating harmonic series>, with sum $-\ln2$. For the third series, the convergent <p-series> $\sum_{n\ge1}1/n^2$ 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.
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