Let $\sum a_n$ be a conditionally convergent sum of complex numbers. Can $\sum a_n z^n$ converge $\forall |z|=1$? I'm fairly new to complex analysis, and I just thought of this problem, but I can't seem to find an easy proof, or an easy counterexample.
 A: Yes, it can.
Let $a_n$ be the sequence 
$$1,-\frac{1}{4},-\frac{1}{4},\frac{1}{9},\frac{1}{9},\frac{1}{9},-\frac{1}{16},-\frac{1}{16},-\frac{1}{16},-\frac{1}{16},\dots$$
Then $\sum |a_n|$ diverges, because the harmonic series diverges.
However, $\sum a_n$ does converge (to $\ln 2$). To see this, note that the partial sums at the end of each block of identical numbers are just the partial sums of the alternating harmonic series. So  we have a subsequence of the sequence of partial sums which converges; moreover, any partial sum of the $a_n$ between the $k$th and $(k+1)$st element of this subsequence differs from the $k$th partial sum in the subsequence by at most $\frac{1}{k}$. So the entire sequence of partial sums converges.
If $|z|=1$ and $z \neq 1$, the sum of the $k$th group of identical terms in $\sum z^n a_n$ will be of the form
$$
\frac{(-1)^{k+1}}{k^2}z^N\frac{1-z^k}{1-z}
$$
for some integer $N$ that's too annoying to compute. This is bounded in absolute value by $$\frac{2}{|1-z|} \frac{1}{k^2}$$
the sum of which converges. So, as above, we have a subsequence of the sequence of partial sums which converges; moreover, any partial sum between the $k$th and $(k+1)$st element of this subsequence again differs from the $k$th partial sum in the subsequence by at most $\frac{1}{k}$. So the series $\sum z^n a_n$ also converges when $z \neq 1$.
