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#1
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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 |
#2
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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. |
#3
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thank you
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#4
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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 |
#5
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yah just take the derivative of the analagous function f(x)
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