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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  
Global Moderator Joined: May 2007 Posts: 6,759 Thanks: 696  Re: Convergence/Divergence of an Infinite Series Quote:
 
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 pseries 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: 6,759 Thanks: 696  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.


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convergence or divergence, infinite, series 
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