Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 9D (Analysis I)

Let $M$ bound all $|a_n{w}^n|$ and put $q=|z|/|w|<1$. ThenThe geometric series ${\sum }_{n=0}^{\infty }M{q}^n$ converges, so the comparison test for series proves absolute convergence of $\sum a_n{z}^n$. This is the bounded power-series terms force interior absolute convergence principle.

Define the radius of convergence byallowing $\rho =0$ and $\rho =\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives absolute convergence at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.

For $z\ne 0$,If $R|z|<1$, choose $q$ strictly between that limit and $1$. Eventually $|a_n{z}^n|\le q^n$, proving absolute convergence by the geometric series comparison. If $R|z|>1$, choose $q>1$ below the limit; eventually $|a_n{z}^n|\ge q^n$, so the terms do not tend to zero and the series diverges. The point $z=0$ is harmless. This proves the root test in the needed case and gives

All three examples below have radius of convergence one, because the $n$th roots of their coefficient moduli tend to one.

For convergence everywhere on the boundary, take ${\sum }_{n=1}^{\infty }{z}^n/n^2$. On $|z|=1$ it converges absolutely: for $n\ge 2$, $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.

For divergence everywhere on the boundary, take ${\sum }_{n=0}^{\infty }{z}^n$. If $|z|=1$, every term has modulus one, so the term test for divergence applies.

For mixed boundary behavior, take ${\sum }_{n=1}^{\infty }{z}^n/n$. At $z=1$ it is the divergent harmonic series; each successive dyadic block contributes at least $1/2$. At every other point $|z|=1$, summation by parts proves convergence. Indeed put $B_k={\sum }_{j=m}^k{z}^j$, so the finite geometric series gives $|B_k|\le 2/|1-z|=:C_z$. ThenThus the partial sums form a Cauchy sequence and converge. This derives the relevant Dirichlet test rather than only naming it.