112 30746 Suppose a 1 1 and 1 2 cos n n n a a n Does \u03a3 a n converge absolutely

112 30746 suppose a 1 1 and 1 2 cos n n n a a n does

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30/746. Suppose a 1 = 1 and 1 2 cos n n n a a n + + = . Does Σ a n converge absolutely, converge conditionally, or diverge? Use ratio test: 2 cos 2 cos lim lim 0 1 n n n n n a n n a n ￲ ￲ ￲ ￲ + + = = < Therefore, the series Σ a n converges absolutely by the ratio test. Root Test Theorem: Suppose lim n n n a L ￲ ￲ = , a n ≠ 0 n , L ≥ 0. i. If L < 1, then Σ a n is absolutely convergent. ii. If L > 1, then Σ a n is divergent. iii. If L = 1, the ratio test gives no information. (Note: Often works for series where | a n | has the form f ( n ) g ( n ) ). 23/746. 2 2 1 1 2 1 n n n n = + + Use root test: 2 2 2 2 1 2 1 1 lim lim 2 1 2 1 1 n n n n n n n n ￲ ￲ ￲ ￲ + + = + + = < Therefore, the series 2 2 1 1 2 1 n n n n = + + converges by the root test. Rearrangements Definition: A rearrangement of Σ a n has the same terms as this series, but in a different order. Formally, Σ a σ(n) is a rearrangement of Σ a n iff σ: is a bijection. Theorem: 1. If Σ a n is absolutely convergent, then every rearrangement is also absolutely convergent, and has the same sum. 2. If Σ a n is conditionally convergent and R (–∞, ∞), then there exist a rearrangement of Σ a n | Σ a σ(n) = R , and there exist rearrangements of Σ a n that do not converge. 113
Example : 1 1 1 1 2 3 4 1 1 ( 1) 1 ln 2 n n n a n - = = - = - + - + = L This series is conditionally convergent 1 1 n n = diverges. Consider ( ) } ( ) } ( ) } 3 5 1 6 2 4 1 1 1 1 1 1 1 1 3 2 5 7 4 9 11 8 1 b b b b b b n b = + - + + - + + - + 678 67 8 6 7 8 L . This series converges by the alternating series convergence test | b n +1 | < | b n | n and lim 0 n n b ￲ ￲ = . 1 1 1 1 1 1 1 2 3 4 5 6 7 8 1 1 1 1 1 1 2 2 4 6 8 2 3 3 1 1 1 1 1 2 3 2 5 7 4 2 1 ln 2 + ln 2 1 0 0 ln 2 n n n n a a a b = - + - + - + - + = = - - + = = + + - + + + - + = = L L L Consider ( ) } ( ) } ( ) } 3 5 1 6 2 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 5 7 2 9 11 13 15 4 17 19 21 23 8 1 c c c c c c n c = + + + - + + + + - + + + + - + 6 447 4 48 6 4 47 4 48 6 4 4 7 4 4 8 L . This series also converges by the alternating convergence test | c n +1 | < | c n | n and lim 0 n n c ￲ ￲ = . It may be shown that Σ c n = 2 ln 2 by a method similar to that given above. 114
11.7 Strategies for Testing 11.7 Strategies for Testing Series for Convergence Series for Convergence 22 March, 2006 1-38/748 recognize geometric and p -series geometric series convergent iff | r | < 1 p -series convergent iff p > 1 recognize series similar to geometric or p -series try comparison or limit comparison tests if lim 0 n n a ￲ ￲ , use divergence test if series is alternating, try alternating series test try ratio test if terms involve exponentials or factorials try root test if terms are of the form f ( n ) g ( n ) try integral test if 1 ( ) 0, ( ) 0, ( ) n a f n f x f x dx = < . 2/748. 2 1 1 n n n n = - + Use limit comparison test with harmonic series (divergent): 2 2 2 2 1 1 lim lim 1 lim 1 1 lim 1 1 1 n n n n n n n n n n n n n n n n n n ￲ ￲ ￲ ￲ ￲ ￲ ￲ ￲ - - + = + - = + - = + = Therefore, this series diverges by the limit comparison test.

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