tag:blogger.com,1999:blog-6629955541128823275.post6434906184242845802..comments2024-03-08T01:15:41.672-05:00Comments on Andrew Eckford: The Blog: A quick exercise on divergent sequencesAndrew Eckfordhttp://www.blogger.com/profile/07739059406915664466noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6629955541128823275.post-63110155394984627392011-05-09T14:11:56.005-04:002011-05-09T14:11:56.005-04:00s(j) = sin(log(j))
This sequence satisfies both c...s(j) = sin(log(j))<br /><br />This sequence satisfies both conditions: -1 <= sin(log(j)) <= 1, and using trig identities it is not too hard to show the second condition. However, it clearly has no limit, as it never stops bouncing between -1 and 1 (though each bounce is on an ever-increasing "period").Andrew Eckfordhttps://www.blogger.com/profile/07739059406915664466noreply@blogger.com