<<

Lecture 24Section 11.4 Absolute and Conditional

Convergence; Alternating

Jiwen He

Basic Series that Converge or Diverge Basic Series that Converge X : 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

Convergence Tests (1) Basic Test for Convergence P Keep in Mind that, if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test. P k P k k Examples 1. x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. [1ex] k X k X  1  diverges since k → 1 6= 0. [1ex] 1 − diverges since k + 1 k+1 k 1 k −1 ak = 1 − k → e 6= 0.

Convergence Tests (2) Comparison Tests Rational terms are most easily handled by basic comparison or comparison with p-series P 1/kp

Basic Comparison Test

1 X 1 X 1 X k3 converges by comparison with converges 2k3 + 1 k3 k5 + 4k4 + 7 X 1 X 1 X 2 by comparison with converges by comparison with k2 k3 − k2 k3 X 1 X 1 X 1 diverges by comparison with diverges by 3k + 1 3(k + 1) ln(k + 6) X 1 comparison with k + 6

Limit Comparison Test X 1 X 1 X 3k2 + 2k + 1 converges by comparison with . diverges k3 − 1 √ k3 k3 + 1 X 3 X 5 k + 100 by comparison with √ √ converges by comparison with k 2k2 k − 9 k X 5 2k2

Convergence Tests (3) and The root test is used only if powers are involved.

Root Test 2 X k 1/k 2 X 1 1/k converges:(a ) = 1 · k1/k → 1 · 1 converges:(a ) = 2k k 2 2 (ln k)k k 2  k k X 1 1/k   1 → 0 1 − converges:(a ) = 1 + (−1) → e−1 ln k k k k

Convergence Tests (4) Root Test and Ratio Test The ratio test is eﬀective with and with combinations of powers and factorials. Ratio Comparison Test 2 2 X k ak+1 1 (k+1) 1 X 1 ak+1 1 converges: = · 2 → converges: = → 0 2k ak 2 k 2 k! ak k+1 k X k a X k a k converges: k+1 = 1 · k+1 → 1 diverges: k+1 = 1 + 1  → e 10k ak 10 k 10 k! ak k k k X 2 ak+1 1−(2/3) 1 X 1 ak+1 converges: = 2 · k → 2 · √ converges: = 3k − 2k ak 3−2(2/3) 3 k! ak q 1 k+1 → 0

2 2.1 Absolute Convergence Absolute Convergence

2 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 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

Conditional Convergence P P P A series ak is said to converge conditionally if ak converges while |ak| diverges.

Alternating p-Series with 0 < p ≤ 1 X (−1)k X 1 , 0 < p ≤ 1, converge conditionally because diverges. ⇒ kp kp ∞ X (−1)k+1 1 1 1 = 1 − + − − · · · converge conditionally. k 2 3 4 k=1 3

Alternating Series Alternating Series Let {ak} be a of positive numbers. X k (−1) ak = a0 − a1 + a2 − a3 + a4 − · · · is called an alternating series. 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 , p > 0, converge since f(x) = is decreasing, i.e., f 0(x) = − > kp xp xp+1 ∞ X (−1)k+1 1 1 1 0 for ∀x > 0, and lim f(x) = 0. ⇒ = 1 − + − − · · · x→∞ k 2 3 4 k=1 converge conditionally.

3 Examples X (−1)k 1 2 , converge since f(x) = is decreasing, i.e., f 0(x) = − > 2k + 1 2x + 1 (2x + 1)2 0 for ∀x > 0, and lim f(x) = 0. x→∞

k 2 X (−1) k x 0 x − 10 2 , converge since f(x) = 2 is decreasing, i.e., f (x) = − 2 2 > k + 10√ x + 10 (x + 10) 0, for ∀x > 10, and lim f(x) = 0. x→∞

An Estimate for Alternating Series

An Estimate for Alternating Series Let {ak} be a decreasing sequence of positive numbers that tends to 0 and let ∞ X k L = (−1) ak. Then the sum L lies between consecutive partial sums sn, k=0 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. 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 = . 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

4 Example ∞ X (−1)k+1 1 1 Find s to approximate = 1 − + ··· within 10−2. n (2k + 1)! 3! 5! k=0 −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 4 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 2. All rearrangements of an absolutely convergent series converge absolutely to the same sum.

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

Remark

5 • 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.

Outline

Contents

1 Convergence Tests 1

2 Absolute Convergence 2 2.1 Absolute Convergence ...... 2

3 Alternating Series 3

4 Rearrangements 5

6