= Solution
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 \b[$\sum_{n=1}^\infty z^n/n^2$]. On $|z|=1$ it converges absolutely: for $n\ge2$, $1/n^2\le1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.
For divergence everywhere on the boundary, take \b[$\sum_{n=0}^\infty z^n$]. If $|z|=1$, every term has modulus one, so the <term test for divergence> applies.
For mixed boundary behavior, take \b[$\sum_{n=1}^\infty z^n/n$]. 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 $B_k=\sum_{j=m}^kz^j$, so the finite <geometric series> gives $|B_k|\le2/|1-z|=:C_z$. Then
$$
\sum_{n=m}^N\frac{z^n}{n}=\frac{B_N}{N}+\sum_{n=m}^{N-1}B_n\left(\frac1n-\frac1{n+1}\right),\qquad
\left|\sum_{n=m}^N\frac{z^n}{n}\right|\le\frac{C_z}{m}.
$$
Thus the partial sums form a <Cauchy sequence> and converge. This derives the relevant <Dirichlet test> rather than only naming it.
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