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: 20,751 Thanks: 2135 
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: 20,751 Thanks: 2135 
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)<an and n___0 an=0 it converges 
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/n1/(6n^3)+o(n^4)))=log(11/(6n^2)+o(n^3))~1/6n^2 which is the general term of a convergent series.


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convergence, series, test 
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