Section 11.5
- 1. Chapter 11: Sequences, Series, and Power Series Section 11.5: Alternating Series and Absolute Convergence Alea Wittig SUNY Albany
- 2. Outline Alternating Series Absolute Convergence and Conditional Convergence
- 3. ▶ So far we have looked at methods to deal with series whose terms are positive. ▶ Here we deal with series where terms are not necessarily positive.
- 5. Alternating Series An alternating series is one whose terms are alternately positive and negative. 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + . . . = X n=1 (−1)n−1 1 n − 1 2 + 2 3 − 3 4 + 4 5 − 5 6 + 6 7 − . . . = X n=1 (−1)n n n + 1 In these examples we have an = (−1)n bn or an = (−1)n−1 bn where bn is a positive number. (bn = |an|.)
- 6. Alternating Series Test Theorem: If the alternating series ∞ X n=1 (−1)n−1 bn = b1 − b2 + b3 − b4 + . . . (bn > 0) satisfies the conditions (i) bn+1 ≤ bn for all n, ie, bn is decreasing. (ii) lim n→∞ bn = 0 then the series is convergent
- 8. Remarks ▶ P an = P (−1)n−1bn or P (−1)nbn where bn > 0, means bn = |an| ▶ If (ii) is false then P an diverges: lim n→∞ |an| ̸= 0 =⇒ lim n→∞ an ̸= 0 by Theorem 6 of the 11.1 of the lecture notes The Test for Divergence: lim n→∞ an ̸= 0 =⇒ ∞ X n=1 an diverges.
- 9. Example 1 1/2 Determine whether the alternating harmonic series X n=1 (−1)n−1 n converges.
- 10. Example 1 2/2 (i) bn+1 ≤ bn: bn = 1 n bn+1 ≤ bn ⇐⇒ 1 n + 1 ≤ 1 n ✓ (ii) limn→∞ bn = 0: lim n→∞ bn = lim n→∞ 1 n = 0✓ Thus the alternating harmonic series converges by AST.
- 11. Example 2 1/2 Determine whether the series ∞ X n=1 (−1)n 3n 4n − 1 converges.
- 12. Example 2 2/2 ∞ X n=1 (−1)n 3n 4n − 1 is alternating with bn = 3n 4n−1 (i) bn+1 ≤ bn 3n + 3 4n − 3 ≤ 3n 4n − 1 ⇐⇒ (3n + 3)(4n − 1) ≤ 3n(4n − 3) ⇐⇒ 12n2 − 3n + 12n − 3 ≤ 12n2 − 9n ⇐⇒ 12n2 − 9n − 3 ≤ 12n2 − 9n✓ (ii) lim n→∞ bn = 0: lim n→∞ 3n 4n − 1 = 3 4 ̸= 0 × Fails (ii) so, as explained in remarks, by the test for divergence the series diverges.
- 13. Example 3 1/3 Test the series ∞ X n=1 (−1)n+1 n2 n3 + 1 for convergence or divergence.
- 14. Example 3 2/3 bn = n2 n3 + 1 (i) We can show that bn is decreasing by showing that f (x) = x2 x3+1 is decreasing by taking the derivative: f ′ (x) = (x3 + 1)(2x) − x2(3x2) (x3 + 1)2 < 0 ⇐⇒ (x3 + 1)(2x) − x2 (3x2 ) < 0 ⇐⇒ 2x4 + 2x − 3x4 < 0 ⇐⇒ 2x − x4 < 0 ⇐⇒ 2 < x3 So bn is decreasing for n ≥ 2 ✓. (ii) lim n→∞ n2 n3 + 1 = 0✓ So the series is convergent by AST.
- 15. Absolute Convergence and Conditional Convergence
- 16. ▶ Given any series P an, we can consider the corresponding series ∞ X n=1 |an| = |a1| + |a2| + |a3| + . . . eg P an = P (−3)n =⇒ P |an| = P 3n eg P an = P (−1)n n =⇒ P an = P 1 n
- 17. Absolute Convergence Definition: A series P an is called absolutely convergent if the series of absolute values P |an| is convergent. Theorem: If a series P an is absolutely convergent, then it is convergent.
- 18. Example 4 1/2 Determine whether the series ∞ X n=1 (−1)n−1 n2 = 1 − 1 22 + 1 32 − 1 42 + . . . converges.
- 19. Example 4 2/2 X |an| = X 1 n2 is a p series with p = 2 > 1 which converges =⇒ X an = X (−1)n−1 n2 is absolutely convergent thus convergent
- 20. Example 5 1/2 Determine whether the series ∞ X n=1 cos n n2 = cos (1) 12 + cos (2) 22 + . . . is convergent or divergent
- 21. Example 5 2/2 ▶ P cos n n2 is an example a series which is positive for some values of n and negative for others, but is not of the form P (−1)nbn. ▶ We use the fact that −1 ≤ cos θ ≤ 1 for all θ to determine if the series is absolutely convergent. 0 ≤ |an| = | cos n| n2 ≤ 1 n2 ▶ Since P 1 n2 is convergent, P |an| is convergent by direct comparison. ▶ Thus P an = P cos n n2 is absolutely convergent, and thus convergent.
- 22. Conditional Convergence Definition A series P an is called conditionally convergent if it is convergent but not absolutely convergent; that is, P an converges but P |an| diverges. eg, The alternating harmonic series is conditionally convergent. ∞ X n=1 (−1)n n converges by AST whereas ∞ X n=1 1 n diverges
- 23. Example 6 1/4 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. (a) ∞ X n=1 (−1)n n3 (b) ∞ X n=1 (−1)n 3 √ n (c) ∞ X n=1 (−1)n n 2n + 1
- 24. Example 6 2/4 (a) ∞ X n=1 (−1)n n3 |an| = 1 n3 =⇒ ∞ X n=1 |an| = ∞ X n=1 1 n3 ← convergent p series p = 3 =⇒ ∞ X n=1 (−1)n n3 is absolutely convergent.
- 25. Example 6 3/4 (b) ∞ X n=1 (−1)n 3 √ n X |an| = X 1 3 √ n is a divergent p-series with p = 1 3 ≤ 1 ▶ This means that X an is not absolutely convergent. ▶ Check for conditional convergence using AST: (i) lim n→∞ bn = lim n→∞ 1 3 √ n = 0 ✓ (ii) bn+1 ≤ bn ⇐⇒ 1 3 √ n + 1 ≤ 1 3 √ n ⇐⇒ 1 ≤ 3 r n + 1 n ⇐⇒ 1 ≤ n + 1 n = 1 + 1 n ✓ So ∞ X n=1 (−1)n 3 √ n is conditionally convergent.
- 26. Example 6 4/4 (c) ∞ X n=1 (−1)n n 2n + 1 lim n→∞ bn = lim n→∞ n 2n + 1 = 1 2 ̸= 0 =⇒ ∞ X n=1 (−1)n n 2n + 1 is divergent by test for divergence.