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  #1  
Old 11-21-2005, 07:44 PM
lotus776 lotus776 is offline
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Default infinite series (alternating series test)...need help

Part of testing for divergence/convergence involves the use of the "Alternating Series Test". I need to prove that a series is both DECREASING and the limit -> 0. I obviously know how to test the limit (which I usually do first) but determining if the series is decresing or not is harder to prove. The simple answer is to look at the denominator and if it "larger" than the numerator than the series is decreasing but my teacher says that is not a sufficient proof.
Please explain how to prove that a series is decreasing

thanks
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  #2  
Old 11-21-2005, 09:59 PM
jason_t jason_t is offline
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Default Re: infinite series (alternating series test)...need help

Here's an example.

Show that sum (-1)^n n/(n^2+1) n = 1 to infinity converges.

Proof: This is an alternating series. Set a_n = n/(n^2+1). Then a_n --> 0.

Set f(x) = x/(x^2+1). Then f'(x) = (1-x^2)/(x^2+1)^2 which is < 0 for x > 1. Therefore f is a decreasing function and this shows that a_n decreases too.

Now apply the alternating series convergence theorem.
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Old 11-22-2005, 09:58 PM
lotus776 lotus776 is offline
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Default Re: infinite series (alternating series test)...need help

thank you
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  #4  
Old 11-24-2005, 04:47 AM
sirio11 sirio11 is offline
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Default Re: infinite series (alternating series test)...need help

Usually simple algebra and mathematical induction will do it.

Using Jason's example

You want to compare

a(n) and a(n+1)

[n/(n^2+1)] and [(n+1)/((n+1)^2+1)]

cross multiply

n[((n+1)^2+1] and (n+1)(n^2+1)

n^3+2n^2+n+1 and n^3+n^2+n+1

n^2 and 0 which is obviously >

Go back and then a(n) > a(n+1) therefore decreasing
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  #5  
Old 11-24-2005, 05:08 AM
twankerr twankerr is offline
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Default Re: infinite series (alternating series test)...need help

yah just take the derivative of the analagous function f(x)
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