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
*** QuickLaTeX cannot compile formula: \sin(x) *** Error message: Cannot connect to QuickLaTeX server: cURL error 60: SSL certificate problem: unable to get local issuer certificate Please make sure your server/PHP settings allow HTTP requests to external resources ("allow_url_fopen", etc.) These links might help in finding solution: http://wordpress.org/extend/plugins/core-control/ http://wordpress.org/support/topic/an-unexpected-http-error-occurred-during-the-api-request-on-wordpress-3?replies=37To 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