5.5 Alternating Series

Transcription

1 5.5 Alternating Series Most of the tests we ve dealt with used a series with positive terms. Now we ll focus on the type of series where the sequence a n alternates between positive and negative terms. Definition 5.0 (Alternating Series). A series whose terms are alternately positive and negative. For example, = ( ) n n=0 ( ) n 2 = n Theorem 5.4 (Alternating Series Test). If the alternating series ( ) n b n = b b 2 + b b 4 + b 5... satisfies the following requirements. b n > 0 2. b n+ b n (terms are decreasing). lim n b n = 0 then the alternating series is convergent. Additionally, we also have the following conclusion If ( ) n b n = S, then 0 < S < a and S 2N < S < S 2N+ 260

2 This conclusion states that if the altnerating series converges, then its sum will always be between consectuve partial sums. Let s take a look at an example. Example Consider the following alternating series ( ) n 2. Note, this is a geometric serires with r = 2, so we know it will converge. It will converge to ( ) n 2 = 2. Let s look at it now as an alternating series ( 2 ) = 5 = 0.6 where b n = test. ( 2 ( ) n ) n ( ) n 2. Note that b n satisfies the requirements for the alternating series (a) b n > 0 (b) b n is decreasing (c) b n 0 as n By the alternating series test, that. ( ) n ( 2 ) n converges. But we already knew. The last part of the alternating series test states a way to estimate the sum by looking at two consecutive partial sums. It states that the true sum of the series will always be 26

3 between two consecutive partial sums. Let s check that. Recall the true sum is S = 0.6. (a) S = Note that we satisfy 0 < S < (b) S 2 = 2 = and < S < (c) S = = and < S < 7 9 (d) S 4 = = 27 = and 8 27 < S < If we keep this process up, we will eventually trap the true sum between two very close numbers. For example, S S or 262

4 S S Example Recall that the harmonic series n diverges (p-series test). What about ( ) n This series is an alternating series. Let b n = n. n. b n = n > 0 2. b n = n is decreasing. b n 0 as n Since it satisfies the altnerating series test, we conclude that Example Determine if n=2 ( ) n converges n ( ) n (n ) 2n + converges. It s clear it altnernates because of the ( ) n. So now let s consider b n = n 2n + if it satisfies the rest of the alternating series test. and see. b n = n 2n + > 0 2. If we want to check to see if b n is decreasing, we need to find the derivative and show it s negative. The problem is the derivative is 26

5 So b n is not decreasing.. Also, b n > 0 for all n (2n + ) 2 The fact that the series is not decreasing to 0 means this series diverges. Recall from a few sections ago that if the sequence a n doesn t converge to 0 means the series a n automatically diverges. Let s return to estimating an alternating series. Theorem 5.5. Let S = ( ) n b n, where b n > 0, decreases, and b n 0. Then R n = S S N < b N+ This theorem states that you can add up the first N terms. The remainder (the stuff that we haven t added yet) must be less than the next term in the sequence b N+. Example: Recall the series ( ) n ( 2 ) n, which converges to S = 0.6. R 0 = S S 0 < b = ( ) 2 = Recall from a couple of examples ago that S 0 = So did our theorem work? so R 0 = =

6 R 0 = < b = Yes, our theorem worked. Let s check another one. so R 20 = = b 2 = R 20 = < b 2 = So what can we get from this method? If we want to estimate to within say, 0.000, we need to find out when R N < or instead b N+ < Example: Consider the following series n=0 ( ) n 0 n n! Estimate the series so it has an error less than In order to use our theorem, it needs to satisfiy the altnerating series test. Let b n = 0 n n!. 265

7 (a) b n = 0 n n! > 0 (b) b n is decreasing. We don t have a derivative to check, but c mon! I think we can agree this decreases. (c) b n 0 as n 2. Now that we know it converges by the Alternating Series Test, let s work on estimating. Initially we want to find N such that R N < But since R N < b N+, it s much easier to find N where b N+ < So let s get started. 0 N+ (N + )! < < 0 N+ (N + )! 266

8 . We just need to check a few values for N (a) If N =, 0 4 4! = < (b) If N = 4, 0 5 5! = > We conclude we need N 4 to get our series within Since n starts at n =, we need the first N = 4 terms to get our required accuracy. Let s verify that we did in fact get our required accuracy. Assume the true sum is S = (a) S 2 = S 2 S = > (b) S = S S = >

9 (c) S 4 = S 4 S = < So it s confirmed. We reach our required accuracy at N =