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 February 26th, 2010, 11:16 PM #1 Member   Joined: Feb 2010 Posts: 31 Thanks: 0 Convergence/Divergence of an Infinite Series I need to know what test to apply to test for convergence of sum from 1 to infinity of [arctan(n)/(n^2)] please help! February 27th, 2010, 01:16 PM   #2
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Re: Convergence/Divergence of an Infinite Series

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 Originally Posted by everettjsj2 I need to know what test to apply to test for convergence of sum from 1 to infinity of [arctan(n)/(n^2)] please help!
Since arctan is multivalued I'll assume that you wish to confine it to the principal branch which is bounded by ?/2. In which case, you can test against the series 1/n^2 February 27th, 2010, 01:58 PM #3 Member   Joined: Feb 2010 Posts: 31 Thanks: 0 Re: Convergence/Divergence of an Infinite Series I tried the Limit Comparison test with a(n) = arctan(n)/n^2 and b(n) = 1/n^2 (convergent p-series with p=2>1) so lim n--> infinity [a(n)/(b(n)] = lim n--> infinity [arctan(n)/n^2]*[n^2/1] = lim n--> infinity arctan(n) = pi/2 hence, lim n--> infinity a(n)/b(n) = pi/2 > 0 both converge or diverge by this property. We know that b(n) converges, therefore a(n) also converges Series is convergent how does that look mathman? February 28th, 2010, 04:36 PM #4 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 723 Re: Convergence/Divergence of an Infinite Series It looks right, but clumsy. All you need is |a(n)|?(?/2)b(n). So you have dominant convergence. Tags convergence or divergence, infinite, series ,

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# is arctan convergent or divergent if it is multivalued

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