User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 March 2nd, 2009, 12:10 PM #1 Newbie   Joined: Mar 2009 Posts: 8 Thanks: 0 Confused I am given a_n > 0, lim (n*a_n) = l, l != 0... show that \Sum (a_n) is divergent. I would like a hint where to start. I thought that since the limit n*a_n = l, that means that n*a_n is bounded by some value greater than 0... but I am unsure if that can help me show that a_n is unbounded. Or am I completely in the wrong area here. Thank you March 2nd, 2009, 01:35 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Confused If and then so by the comparison of limits test, converges iff converges. March 2nd, 2009, 03:52 PM #3 Newbie   Joined: Mar 2009 Posts: 8 Thanks: 0 Re: Confused Thanks... The limit comparison test isn't a theorem we can use, but it gave me an idea on how to do the problem if a_n >0 for all and and lim (n*a_n) = l, that implies that n*a_n is bounded. Thus there exist real numbers c,d such that c <= n*a_n <= d <=> 1/n * c <= a_n. Since Sum (1/n) diverges (sum c(1/n) diverges) then by the comparison test Sum(a_n) must diverge. by similar logic if lim n^2*a_n, converges, a_n > 0 then Sum (a_n) must converge. Tags confused Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post charlieboon Calculus 5 February 23rd, 2013 10:01 AM ccfoose Algebra 3 April 16th, 2012 09:55 PM Siedas Algebra 9 March 10th, 2012 01:29 PM momoftwo New Users 4 August 19th, 2009 06:15 PM charlieboon Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      