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 February 19th, 2007, 05:20 PM #1 Newbie   Joined: Dec 2006 Posts: 5 Thanks: 0 the series for convergence test # sum # log(n.(sin1/n)) is this series convergent or divergent and how???
 February 19th, 2007, 06:52 PM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Are you asking whether: ∑(n=1, n=∞) log(n*sin(1/n)) converges or diverges?
 February 20th, 2007, 01:28 AM #3 Newbie   Joined: Dec 2006 Posts: 5 Thanks: 0 yes i couldn't write it like that.That's it
 February 20th, 2007, 09:01 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,389 Thanks: 2015 It converges.
 February 23rd, 2007, 01:23 PM #5 Newbie   Joined: Dec 2006 Posts: 5 Thanks: 0 i know that it is convergent.I want just a solution
 February 24th, 2007, 03:15 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,389 Thanks: 2015 Approximate each term as a fixed multiple of a power of n.
 May 25th, 2007, 10:08 PM #7 Newbie   Joined: May 2007 From: artesia,ca Posts: 2 Thanks: 0 it converges ∑(n=1, n=∞) log(n*sin(1/n)) converges let me show u lim an=n___∞ log(n*sin(1/n)) using l'hopital rule lim an=n___∞ log((sin(1/n)/(1/n))) lim lim an=n___∞ log((1))=0here as a(n+1)
 May 26th, 2007, 12:02 AM #8 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 log(n*sin(1/n)) is equivalent to log(n*(1/n-1/(6n^3)+o(n^4)))=log(1-1/(6n^2)+o(n^3))~-1/6n^2 which is the general term of a convergent series.

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