Solution
= Solution
Taking real parts of the geometric sum gives
$$
a_n=\Re\frac{1-e^{i(n+1)y}}{1-e^{iy}},
$$
so $(a_n)$ is bounded when $\cos y\ne1$. It cannot converge: convergence would imply $\cos(ny)=a_n-a_{n-1}\to0$, but then
$$
\cos(2ny)=2\cos^2(ny)-1\to-1,
$$
whereas the same necessary condition applied to the subsequence $2n$ would give a <limit> of zero.