= Solution
The key fact is that convergence of a <power series> at one nonzero point forces <absolute convergence> at every smaller modulus. Indeed, if $\sum a_nw^n$ converges and $w\ne0$, its terms form a <bounded sequence>, say $|a_nw^n|\le M$. For $|z|<|w|$,
$$
\sum_{n=0}^{\infty}|a_nz^n|\le M\sum_{n=0}^{\infty}\left|\frac zw\right|^n<\infty
$$
by convergence of a <geometric series>.
Let $S$ be the set of moduli of points at which the <power series> converges, and put $R=\sup S$, allowing $R=\infty$. The set contains zero, so its <supremum> is defined in $[0,\infty]$. Whenever $|z|<R$, the definition of <supremum> provides a convergence point $w$ with $|w|>|z|$. The preceding estimate gives <absolute convergence> at $z$. Whenever $|z|>R$, convergence at $z$ would put $|z|$ in $S$, contradicting its upper bound. This proves the asserted <radius of convergence> property, including $R=0$ and $R=\infty$. Nothing in the argument decides convergence on $|z|=R$.
For the requested boundary behavior, take the <power series>
$$
\boxed{\sum_{k=1}^{\infty}\frac{z^{2k}}k.}
$$
Its coefficients are $a_{2k}=1/k$ for $k\ge1$, $a_{2k+1}=0$ and $a_0=0$. At either $z=1$ or $z=-1$ it is the divergent <harmonic series>. At either $z=i$ or $z=-i$ it is $\sum_{k\ge1}(-1)^k/k$, which converges by the <alternating series test>. In fact its <radius of convergence> is one, since inside the unit circle the terms are bounded by a <geometric series>, and outside it they do not tend to zero.
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