Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 5F / d / ii / Solution

Taking real parts of the geometric sum givesso $(a_n)$ is bounded when $\cos y\ne 1$. It cannot converge: convergence would imply $\cos (ny)=a_n-a_{n-1}\to 0$, but thenwhereas the same necessary condition applied to the subsequence $2n$ would give a limit of zero.

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