Consider the series
![\[ \sum_{n=1}^{\infty} \sin (\log n). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5ee94edb7ca4eba426131e56be2e4b05_l3.png "Rendered by QuickLaTeX.com")
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The given series diverges since
![\[ \lim_{n \to \infty} \sin (\log n) \neq 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e6e83b6b339febe2acf4527114d63e30_l3.png "Rendered by QuickLaTeX.com")
Consider the series
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The given series diverges since
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