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L20 Alternating Series Test

# L20 Alternating Series Test - remainder is R n = s ± s n...

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Lecture 20: Alternating Test The terms of an alternating series alternate in signs: 1 X n =1 ± ± 1 2 ² n = ± 1 = 2 + 1 = 4 ± 1 = 8 + 1 = 16 ± ::: 1 X k =1 ( ± 1) k ± 1 k = 1 ± 1 = 2 + 1 = 3 ± 1 = 4 + 1 = 5 ± ::: So an alternating series P a n has the form a n = ( ± 1) n ± 1 b n or a n = ( ± 1) n b n with j a n j = b n ; b n ² 0 1

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Alternating Series Test{ If the alternating series 1 X n =1 a n = 1 X n =1 ( ± 1) n ± 1 b n = b 1 ± b 2 + :::; b n > 0 satis±es (i) b n +1 ² b n ; and (ii) lim n !1 b n = 0 then the series is convergent. NOTE: You must verify both parts of the alternating series test. ex. 1 X n =1 ( ± 1) n +1 ( n + 1) n (Here, the second condition of the alternating series test is satis±ed. In fact, the series .) 2
ex. 1 X n =1 ( ± 1) n +1 n ex. 1 X n =1 ( ± 1) n 3 n 4 n ± 1 3

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ex. (#12) 1 X n =1 ( ± 1) n ± 1 e 1 =n n (conv.) 4
ex. (#16) 1 X n =1 sin( n±= 2) n ! (conv.) 5

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Estimaing Sums Let s be the sum of a convergent algernating series that satis±es both conditions of the test, and let s n be the n th partial sum. Then the

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Unformatted text preview: remainder is R n = s ± s n : and j R n j = j s ± s n j ² b n +1 : (This rule doesn’t apply to any other types of series.) pf. 6 Show that the series converges. How many terms of the series do we need to add in order to ±nd the sum to the indicated accuracy? ex. (#26) 1 X n =1 ( ± 1) n ± 1 ne ± n ( j error j < : 01) 7 ex. For what values of p is the series conver-gent? (#32) 1 X n =1 ( ± 1) n ± 1 n p NYTI: ex. 1 X n =1 ( ± 1) n n n n ! ex. 1 X n =1 n ( ± 2) n ± 1 (Div, conv.) Students are responsible for problems in 11.5: 2-20, 32-33, 24,25. 8...
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