= Solution
For <absolute convergence>, examine $\sum_{n\geq1}|x-3|^n/n$. When $0<|x-3|<1$, the ratio of successive terms tends to $|x-3|<1$, so the <ratio test> proves convergence; at $x=3$ every term is zero. For $|x-3|>1$, the terms do not tend to zero.
At either boundary $|x-3|=1$, the absolute-value series is the divergent <harmonic series>. Thus for real $x$ the complete answer is
$$
\boxed{|x-3|<1,\qquad 2<x<4.}
$$
The original series at $x=2$ is conditionally convergent by the alternating-series criterion, but it is not absolutely convergent and is correctly excluded here.
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