= Solution
Set $w=e^z$. Then
$$
\sum_{n=0}^\infty b_ne^{nz}
=\sum_{n=0}^\infty b_nw^n.
$$
The hypothesis that convergence and divergence both occur implies that this ordinary power series has a radius $R$ with $0<R<\infty$. Since $|w|=e^{\operatorname{Re}z}$, the <half-plane of convergence of an exponential power series> is determined by
$$
\boxed{S=\log R}:
$$
the series converges for $\operatorname{Re}z<S$ and diverges for $\operatorname{Re}z>S$.
For the specified series,
$$
\frac{2^ne^{inz}}{(n+1)^2}
=\frac{(2e^{iz})^n}{(n+1)^2}.
$$
Writing $z=x+iy$ gives $|2e^{iz}|=2e^{-y}$. The series converges absolutely when $y>\log2$ and diverges by the term test when $y<\log2$. On $y=\log2$, its absolute values are $(n+1)^{-2}$, so it still converges absolutely. Therefore the exact convergence set is
$$
\boxed{\{z\in\mathbb C:\operatorname{Im}z\geq\log2\}}.
$$
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