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February 26th, 2010, 11:16 PM   #1
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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!
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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
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February 27th, 2010, 01:58 PM   #3
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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?
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February 28th, 2010, 04:36 PM   #4
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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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