10.6 Alternating Series, Absolute and Conditional Convergence
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1 10.6 Alternating Series, Absolute and Conditional Convergence
2 The Theorem Theorem The series converges if: n=1 1 The u n s are all positive 2 u n u n+1 n N, N Z 3 u n 0 ( 1) k+1 u n = u 1 u 2 + u 3
3 First Determine if the following series converges. k=1( 1) k+1 1 k
4 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive?
5 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive? 1 k > 0 k
6 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive? 1 k > 0 k 2 Are the terms non-increasing in the tail?
7 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive? 1 k > 0 k 2 Are the terms non-increasing in the tail? 1 k > 1 k+1 k
8 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive? 1 k > 0 k 2 Are the terms non-increasing in the tail? 1 k > 1 k+1 k 3 Does the n th term converge to 0?
9 First Determine if the following series converges. k=1( 1) k+1 1 k We have 3 criteria to consider. 1 Are the u n s all positive? 1 k > 0 k 2 Are the terms non-increasing in the tail? 1 k > 1 k+1 k 3 Does the n th term converge to 0? 1 k 0 Therefore, the alternating harmonic series converges even though the harmonic series diverges.
10 Second Determine if the following series converges. k=1 ( 1) k+1 k + 3 k(k + 1)
11 Second Determine if the following series converges. k=1 ( 1) k+1 k + 3 k(k + 1) We have 3 criteria to consider.
12 Second 1 Are the u n s all positive?
13 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k
14 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail?
15 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k
16 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is
17 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is k(k + 1)(k + 2). When we multiply both sides by that denominator, we get (k + 2)(k + 3) > k(k + 4)
18 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is k(k + 1)(k + 2). When we multiply both sides by that denominator, we get (k + 2)(k + 3) > k(k + 4) k 2 + 5k + 6 > k 2 + 4k
19 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is k(k + 1)(k + 2). When we multiply both sides by that denominator, we get which is true for all k > 0. (k + 2)(k + 3) > k(k + 4) k 2 + 5k + 6 > k 2 + 4k 2k + 6 > 0
20 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is k(k + 1)(k + 2). When we multiply both sides by that denominator, we get (k + 2)(k + 3) > k(k + 4) k 2 + 5k + 6 > k 2 + 4k which is true for all k > 0. 3 Does the n th term converge to 0? 2k + 6 > 0
21 Second 1 Are the u n s all positive? k+3 k(k+1) > 0 k 2 Are the terms non-increasing in the tail? We need k + 3 k(k + 1) > k + 4 (k + 1)(k + 2) k We need to find a common denominator to decide here to compare and see which is larger. That common denominator is k(k + 1)(k + 2). When we multiply both sides by that denominator, we get (k + 2)(k + 3) > k(k + 4) k 2 + 5k + 6 > k 2 + 4k 2k + 6 > 0 which is true for all k > 0. 3 Does the n th term converge to 0? k+3 k(k+1) 0 Therefore the series converges.
22 Some Terms Definition A series a n converges absolutely (is absolutely convergent) if the corresponding series a n converges.
23 Some Terms Definition A series a n converges absolutely (is absolutely convergent) if the corresponding series a n converges. Definition A series that converges but does not converge absolutely converges conditionally.
24 Defining s The alternating harmonic series converges
25 Defining s The alternating harmonic series converges conditionally.
26 Defining s The alternating harmonic series converges conditionally. The series 1 k 2 converges
27 Defining s The alternating harmonic series converges conditionally. The series 1 k 2 converges absolutely.
28 Another Theorem Theorem If converges then converges. a n k=1 a n k=1
29 Another Theorem Theorem If converges then converges. a n k=1 a n k=1 Note: The converse is not true.
30 One More Determine if the given series converges
31 One More Determine if the given series converges What type of series is this?
32 One More Determine if the given series converges What type of series is this? When we write this using series notation, we see that k=0 1 is a geometric series with r < 1. So, since 1 2 k converges, our series converges absolutely. 2 k
33 Notes Notes Rearranging the terms of the series does not change whether or not the series converges and does not change the limit if the series does converge.
34 Notes Notes Rearranging the terms of the series does not change whether or not the series converges and does not change the limit if the series does converge. Deleting a finite number of terms does not change whether or not a series converges but may/will change the limiting value of the series.
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