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Determine the convergence of the series sin (log n)

Consider the series

    \[ \sum_{n=1}^{\infty} \sin (\log n). \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series diverges since

    \[ \lim_{n \to \infty} \sin (\log n) \neq 0. \]

One comment

  1. Anonymous says:

    To show sin(log(n)) doesn’t converge to 0, first assume that it does, and show that this assumption leads to two contradictory results. (1) that cos^2(log(n)) converges to 1, and (2) that cos^2(log(2n)) converges to 0; the latter is done by considering the convergence of (sin(log(4n))-sin(log(n))) using trig and log properties. The problem with this approach though is that it uses the result that every subsequence of a convergent sequence converges to the same limit – which is not taught in Apostol’s Calculus Vol. I. However the convergent subsequence result is discussed in Apostol’s book on analysis. I got help by viewing this video by Dr. Barker: https://www.youtube.com/watch?v=9mo4u_FGvFw&lc=UgztrQGrU7dQyrUMQ0J4AaABAg.9qb9rO7ELl39qcWiFxiOoR

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