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 July 26th, 2014, 12:21 PM #1 Senior Member   Joined: Jul 2014 From: united states Posts: 114 Thanks: 5 convergence/conditional convergence/ divergence determine whether the series is absolutely convergent, conditionally convergent, or divergent. $\sum_{n=1}^{\infty} (\frac{n^2+2}{3n^2+2})^n$ how to do this problem?? im thinking root test. but how would i tell if its convergent or conditionally convergent? July 26th, 2014, 12:51 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus root test $\displaystyle \limsup_{n\to\infty}\sqrt[n]{\left|\frac{n^2+2}{3n^2+2}\right|^{n}}=\limsup_{n \to\infty}\left|\frac{n^2+2}{3n^2+2}\right|=\frac{ 1}{3}<1$, so the series is convergent. Another example for you to consider, take the series $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n}<\infty$ (conditionally convergent), because it is not absolutely convergent $\displaystyle \sum_{n=1}^{\infty}\left|\frac{(-1)^n}{n}\right|=\sum_{n=1}^{\infty}\frac{1}{n}= \infty$ (harmonic series). Thanks from ineedhelpnow Last edited by ZardoZ; July 26th, 2014 at 01:11 PM. July 26th, 2014, 02:38 PM #3 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 General comment: If all the terms of a series are non-negative, then series is either divergent or absolutely convergent. Tags convergence or, divergence 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 razzatazz Real Analysis 8 May 10th, 2013 11:30 PM math221 Calculus 3 April 8th, 2013 06:39 AM FreaKariDunk Real Analysis 28 April 30th, 2012 08:22 PM MathematicallyObtuse Algebra 3 January 29th, 2011 10:35 PM Noworry Calculus 0 December 31st, 1969 04:00 PM

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