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April 28th, 2012, 10:17 PM   #21
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Re: Convergence and Divergence's

We just need for the magnitude of each successive term to decrease at a certain critical rate or beyond to get a convergent series, regardless of how large the first terms are.
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April 29th, 2012, 05:46 PM   #22
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Re: Convergence and Divergence's

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It in fact converges to .

Is this the formula?
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April 29th, 2012, 07:31 PM   #23
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Re: Convergence and Divergence's

No, that formula (I think you mean ) is used in the ratio test, which was inconclusive on the series. I found the sum using the computer...I bet our friend Zardoz could show how to manually find it.
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April 29th, 2012, 07:37 PM   #24
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Re: Convergence and Divergence's

You know about him?

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April 29th, 2012, 07:42 PM   #25
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Re: Convergence and Divergence's

Yes, our resident analyst and I go waaaay back.
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April 29th, 2012, 07:43 PM   #26
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Re: Convergence and Divergence's

It really is a boys club, aye?
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April 29th, 2012, 07:50 PM   #27
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Re: Convergence and Divergence's

Unfortunately...yes. :P
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April 30th, 2012, 08:03 PM   #28
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Re: Convergence and Divergence's

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1.
Will you show me the limit comparison test?

I cannot seem to follow it in the book
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April 30th, 2012, 08:22 PM   #29
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Re: Convergence and Divergence's

By simple inspection we can tell that diverges, and the use of the limit comparison test would really be redundant.

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