 Solution

For absolute convergence, examine ${\sum }_{n\ge 1}|x-3{|}^n/n$. 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 isThe original series at $x=2$ is conditionally convergent by the alternating-series criterion, but it is not absolutely convergent and is correctly excluded here.