Lecture 24 Section 11.4 Absolute and Conditional Convergence; Alternating Series
Convergence Tests Absolute Convergence Alternating Series Rearrangements
Lecture 24 Section 11.4 Absolute and Conditional Convergence; Alternating Series
Jiwen He
Department of Mathematics, University of Houston [email protected] http://math.uh.edu/∼jiwenhe/Math1432
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 1 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Basic Series that Converge or Diverge
Basic Series that Converge X Geometric series: xk , if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 2 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (1)
Basic Test for Convergence Keep in Mind that, P if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test.
Examples P k P k k x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. X k diverges since k → 1 6= 0. k + 1 k+1 k X 1 k 1 − diverges since a = 1 − 1 → e−1 6= 0. k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 3 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Basic Comparison Test X 1 X 1 converges by comparison with 2k3 + 1 k3 X k3 X 1 converges by comparison with k5 + 4k4 + 7 k2 X 1 X 2 converges by comparison with k3 − k2 k3 X 1 X 1 diverges by comparison with 3k + 1 3(k + 1) X 1 X 1 diverges by comparison with ln(k + 6) k + 6
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Limit Comparison Test X 1 X 1 converges by comparison with . k3 − 1 k3 X 3k2 + 2k + 1 X 3 diverges by comparison with √k3 + 1 k X 5 k + 100 X 5 √ √ converges by comparison with 2k2 k − 9 k 2k2
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Limit Comparison Test X 1 X 1 converges by comparison with . k3 − 1 k3 X 3k2 + 2k + 1 X 3 diverges by comparison with √k3 + 1 k X 5 k + 100 X 5 √ √ converges by comparison with 2k2 k − 9 k 2k2
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Limit Comparison Test X 1 X 1 converges by comparison with . k3 − 1 k3 X 3k2 + 2k + 1 X 3 diverges by comparison with √k3 + 1 k X 5 k + 100 X 5 √ √ converges by comparison with 2k2 k − 9 k 2k2
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (2)
Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp
Limit Comparison Test X 1 X 1 converges by comparison with . k3 − 1 k3 X 3k2 + 2k + 1 X 3 diverges by comparison with √k3 + 1 k X 5 k + 100 X 5 √ √ converges by comparison with 2k2 k − 9 k 2k2
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 4 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (3)
Root Test and Ratio Test The root test is used only if powers are involved.
Root Test 2 X k 2 converges: (a )1/k = 1 · k1/k → 1 · 1 2k k 2 2 X 1 converges: (a )1/k = 1 → 0 (ln k)k k ln k 2 k k X 1 (−1) 1 − converges: (a )1/k = 1 + → e−1 k k k
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 5 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with factorials and with combinations of powers and factorials. Ratio Comparison Test
2 2 X k ak+1 1 (k+1) 1 converges: = · 2 → 2k ak 2 k 2 X 1 a converges: k+1 = 1 → 0 k! ak k+1 X k a converges: k+1 = 1 · k+1 → 1 10k ak 10 k 10 k X k a k diverges: k+1 = 1 + 1 → e k! ak k k k X 2 ak+1 1−(2/3) 1 converges: = 2 · k → 2 · 3k − 2k ak 3−2(2/3) 3 X 1 a q √ converges: k+1 = 1 → 0 k! ak k+1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 6 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Absolute Convergence Absolute Convergence P P A series ak is said to converge absolutely if |ak | converges. P P if |ak | converges, then ak converges. i.e., absolutely convergent series are convergent. Alternating p-Series with p > 1 X (−1)k X 1 , p > 1, converge absolutely because converges. kp kp ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge absolutely. k2 22 32 42 k=1 Geometric Series with −1 < x < 1 X X (−1)j(k)xk , −1 < x < 1, converge absolutely because |x|k converges. 1 1 1 1 1 1 ⇒ 1 − − + − + + − · · · converge absolutely. 2 22 23 24 25 26
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 7 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Conditional Convergence
Conditional Convergence P P A series ak is said to converge conditionally if ak converges P while |ak | diverges.
Alternating p-Series with 0 < p ≤ 1 X (−1)k X 1 , 0 < p ≤ 1, converge conditionally because kp kp diverges. ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 8 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Conditional Convergence
Conditional Convergence P P A series ak is said to converge conditionally if ak converges P while |ak | diverges.
Alternating p-Series with 0 < p ≤ 1 X (−1)k X 1 , 0 < p ≤ 1, converge conditionally because kp kp diverges. ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 8 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Conditional Convergence
Conditional Convergence P P A series ak is said to converge conditionally if ak converges P while |ak | diverges.
Alternating p-Series with 0 < p ≤ 1 X (−1)k X 1 , 0 < p ≤ 1, converge conditionally because kp kp diverges. ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 8 / 16 Convergence Tests Absolute Convergence Alternating Series RearrangementsAbsolute Convergence Conditional Convergence
Conditional Convergence P P A series ak is said to converge conditionally if ak converges P while |ak | diverges.
Alternating p-Series with 0 < p ≤ 1 X (−1)k X 1 , 0 < p ≤ 1, converge conditionally because kp kp diverges. ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 8 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Alternating Series Alternating Series
Let {ak } be a sequence of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers. X k If ak → 0, then (−1) ak converges. Alternating p-Series with p > 0 X (−1)k 1 , p > 0, converge since f (x) = is decreasing, i.e., kp xp p f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. xp+1 x→∞ ∞ X (−1)k+1 1 1 1 ⇒ = 1 − + − − · · · converge conditionally. k 2 3 4 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 9 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Examples
X (−1)k 1 , converge since f (x) = is decreasing, i.e., 2k + 1 2x + 1 2 f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. (2x + 1)2 x→∞
X (−1)k k x , converge since f (x) = is decreasing, i.e., k2 + 10 x2 + 10 x2 − 10 √ f 0(x) = − > 0, for ∀x > 10, and lim f (x) = 0. (x2 + 10)2 x→∞
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 10 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Examples
X (−1)k 1 , converge since f (x) = is decreasing, i.e., 2k + 1 2x + 1 2 f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. (2x + 1)2 x→∞
X (−1)k k x , converge since f (x) = is decreasing, i.e., k2 + 10 x2 + 10 x2 − 10 √ f 0(x) = − > 0, for ∀x > 10, and lim f (x) = 0. (x2 + 10)2 x→∞
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 10 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Examples
X (−1)k 1 , converge since f (x) = is decreasing, i.e., 2k + 1 2x + 1 2 f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. (2x + 1)2 x→∞
X (−1)k k x , converge since f (x) = is decreasing, i.e., k2 + 10 x2 + 10 x2 − 10 √ f 0(x) = − > 0, for ∀x > 10, and lim f (x) = 0. (x2 + 10)2 x→∞
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 10 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Examples
X (−1)k 1 , converge since f (x) = is decreasing, i.e., 2k + 1 2x + 1 2 f 0(x) = − > 0 for ∀x > 0, and lim f (x) = 0. (2x + 1)2 x→∞
X (−1)k k x , converge since f (x) = is decreasing, i.e., k2 + 10 x2 + 10 x2 − 10 √ f 0(x) = − > 0, for ∀x > 10, and lim f (x) = 0. (x2 + 10)2 x→∞
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 10 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements An Estimate for Alternating Series
An Estimate for Alternating Series
Let {ak } be a decreasing sequence of positive numbers that tends ∞ X k to 0 and let L = (−1) ak . Then the sum L lies between k=0 consecutive partial sums sn, sn+1,
sn < L < sn+1, if n is odd; sn+1 < L < sn, if n is even.
and thus sn approximates L to within an+1
|L − sn| < an+1.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 11 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements An Estimate for Alternating Series
An Estimate for Alternating Series
Let {ak } be a decreasing sequence of positive numbers that tends ∞ X k to 0 and let L = (−1) ak . Then the sum L lies between k=0 consecutive partial sums sn, sn+1,
sn < L < sn+1, if n is odd; sn+1 < L < sn, if n is even.
and thus sn approximates L to within an+1
|L − sn| < an+1.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 11 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements An Estimate for Alternating Series
An Estimate for Alternating Series
Let {ak } be a decreasing sequence of positive numbers that tends ∞ X k to 0 and let L = (−1) ak . Then the sum L lies between k=0 consecutive partial sums sn, sn+1,
sn < L < sn+1, if n is odd; sn+1 < L < sn, if n is even.
and thus sn approximates L to within an+1
|L − sn| < an+1.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 11 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements An Estimate for Alternating Series
An Estimate for Alternating Series
Let {ak } be a decreasing sequence of positive numbers that tends ∞ X k to 0 and let L = (−1) ak . Then the sum L lies between k=0 consecutive partial sums sn, sn+1,
sn < L < sn+1, if n is odd; sn+1 < L < sn, if n is even.
and thus sn approximates L to within an+1
|L − sn| < an+1.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 11 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n k 2 3 k=1
∞ ∞ X (−1)k+1 X (−1)k Set = . For |L − s | < 10−2, we want k k + 1 n k=1 k=0 1 a = < 10−2 ⇒ n + 2 > 102 ⇒ n > 98. n+1 (n + 1) + 1 Then n = 99 and the 99th partial sum s100 is 1 1 1 1 1 s = 1 − + − + ··· + − ≈ 0.6882. 99 2 3 4 99 100 From the estimate 1 |L − s | < a = ≈ 0.00991. 99 100 101 we conclude that ∞ X (−1)k+1 s ≈ 0.6882 < = ln 2 < 0.6981 ≈ s 99 k 100 k=1 Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 12 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Example
∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within n (2k + 1)! 3! 5! k=0 10−2.
−2 For |L − sn| < 10 , we want 1 a = < 10−2 ⇒ n ≥ 1. n+1 (2(n + 1) + 1)! Then n = 1 and the 2nd partial sum s2 is 1 s = 1 − ≈ 0.8333 1 3! From the estimate 1 |L − s | < a = ≈ 0.0083. 1 2 5! we conclude that ∞ X (−1)k+1 s ≈ 0.8333 < = sin 1 < 0.8416 ≈ s 1 (2k + 1)! 2 k=0
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 13 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· ? = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞ X (−1)k 1 1 1 1 1 2 = 1 − + − + − + ··· = absolutely 2k 2 22 23 24 25 3 k=0 1 1 1 1 1 1 1 1 2 Rearrangement 1 + − + + − + + − ··· = 22 2 24 26 23 28 210 25 3
Theorem All rearrangements of an absolutely convergent series converge absolutely to the same sum.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 14 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· ? = ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2) Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6
1 Multiply the original series by 2 ∞ 1 X (−1)k+1 1 1 1 1 1 1 = − + − + + ··· = ln 2 2 k 2 4 6 8 10 2 k=1 Adding the two series, we get the rearrangement ∞ ∞ X (−1)k+1 1 X (−1)k+1 1 1 1 1 1 3 + = 1+ − + + − +··· = ln 2 k 2 k 3 2 5 7 4 2 k=1 k=1
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2)
Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6 Remark A series that is only conditionally convergent can be rearranged to converge to any number we please. It can also be arranged to diverge to +∞ or −∞, or even to oscillate between any two bounds we choose.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2)
Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6 Remark A series that is only conditionally convergent can be rearranged to converge to any number we please. It can also be arranged to diverge to +∞ or −∞, or even to oscillate between any two bounds we choose.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Why Absolute Convergence Matters: Rearrangements (2)
Rearrangement of Conditional Convergence Series
∞ X (−1)k+1 1 1 1 1 1 = 1 − + − + − + ··· = ln 2 conditionally k 2 3 4 5 6 k=1 1 1 1 1 1 1 1 1 Rearrangement 1 + − + + − + + − ··· 6= ln 2 3 2 5 7 4 9 11 6 Remark A series that is only conditionally convergent can be rearranged to converge to any number we please. It can also be arranged to diverge to +∞ or −∞, or even to oscillate between any two bounds we choose.
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 15 / 16 Convergence Tests Absolute Convergence Alternating Series Rearrangements Outline
Convergence Tests
Absolute Convergence Absolute Convergence
Alternating Series
Rearrangements
Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 24 April 10, 2008 16 / 16