My Math Forum convergence/conditional convergence/ divergence

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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.

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