121616949-math.269 - 10.6 Absolute Convergence 255 P an...

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10.6 Absolute Convergence 255 So given a series a n with both positive and negative terms, you should first ask whether | a n | converges. This may be an easier question to answer, because we have tests that apply specifically to terms with non-negative terms. If | a n | converges then you know that a n converges as well. If | a n | diverges then it still may be true that a n converges—you will have to do more work to decide the question. Another way to think of this result is: it is (potentially) easier for a n to converge than for | a n | to converge, because the latter series cannot take advantage of cancellation. If | a n | converges we say that a n converges absolutely ; to say that a n converges absolutely is to say that any cancellation that happens to come along is not really needed, as the terms already get small so fast that convergence is guaranteed by that alone. If a n converges but | a n | does not, we say that a n converges conditionally . For example X n =1 ( - 1) n - 1
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Unformatted text preview: 2 converges absolutely, while ∞ X n =1 (-1) n-1 1 n converges conditionally. EXAMPLE 10.37 Does ∞ X n =2 sin n n 2 converge? In example 10.32 we saw that ∞ X n =2 | sin n | n 2 converges, so the given series converges abso-lutely. EXAMPLE 10.38 Does ∞ X n =0 (-1) n 3 n + 4 2 n 2 + 3 n + 5 converge? Taking the absolute value, ∞ X n =0 3 n + 4 2 n 2 + 3 n + 5 diverges by comparison to ∞ X n =1 3 10 n , so if the series converges it does so conditionally. It is true that lim n →∞ (3 n + 4) / (2 n 2 + 3 n + 5) = 0, so to apply the alternating series test we need to know whether the terms are decreasing. If we let f ( x ) = (3 x + 4) / (2 x 2 + 3 x + 5) then f ± ( x ) =-(6 x 2 + 16 x-3) / (2 x 2 + 3 x + 5) 2 , and it is not hard to see that this is negative for x ≥ 1, so the series is decreasing and by the alternating series test it converges....
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  • Fall '07
  • JonathanRogawski
  • Math, Calculus, Mathematical Series, |an | converges

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