Absolute Convergence and the Ratio & Root Tests

Transcription

1 Absolute Convergence and the Ratio & Root Tests Math114 Department of Mathematics, University of Kentucky February 20, 2017 Math114 Lecture 15 1/ 12

2 ( 1) n 1 = Math114 Lecture 15 2/ 12

3 ( 1) n 1 = Is the following procedure valid? Math114 Lecture 15 2/ 12

4 ( 1) n 1 = Is the following procedure valid? ( 1) n 1 = (1 1) + (1 1) + (1 1) + (1 1) + = = 0 Math114 Lecture 15 2/ 12

5 ( 1) n 1 = Is the following procedure valid? ( 1) n 1 = (1 1) + (1 1) + (1 1) + (1 1) + = = 0 What about... Math114 Lecture 15 2/ 12

6 ( 1) n 1 = Is the following procedure valid? ( 1) n 1 = (1 1) + (1 1) + (1 1) + (1 1) + = = 0 What about... ( 1) n 1 = 1 + ( 1 + 1) + ( 1 + 1) + ( 1 + 1) + = = 1 Math114 Lecture 15 2/ 12

7 ( 1) n 1 = Is the following procedure valid? ( 1) n 1 = (1 1) + (1 1) + (1 1) + (1 1) + = = 0 What about... ( 1) n 1 = 1 + ( 1 + 1) + ( 1 + 1) + ( 1 + 1) + = = 1 Clearly grouping terms in this manner is NOT valid for this problem. Math114 Lecture 15 2/ 12

8 In fact, {S n } = (1, 0, 1, 0, 1, 0). So, lim n S n DNE. Hence, by the divergence test, ( 1)n 1 diverges. Math114 Lecture 15 3/ 12

9 Consider the Alternating Harmonic Series: ( 1) n 1 1 n = = ln(2). Math114 Lecture 15 4/ 12

10 Consider the Alternating Harmonic Series: ( 1) n 1 1 n = = ln(2). However, = 3 2 ln(2). Math114 Lecture 15 4/ 12

11 When can we treat infinite sums more like finite sums? Math114 Lecture 15 5/ 12

12 Definitions (i) A series a n is called absolutely convergent if the series an converges. (ii) A convergent series that is NOT absolutely convergent is called conditionally convergent. Math114 Lecture 15 6/ 12

13 Definitions (i) A series a n is called absolutely convergent if the series an converges. (ii) A convergent series that is NOT absolutely convergent is called conditionally convergent. Example: The alternating harmonic series is conditionally convergent because ( 1) n 1 1 n converges (by the Alternating Series Test), but diverges (by the Integral Test). ( 1) n 1 1 n = 1 n Math114 Lecture 15 6/ 12

14 Example: ( 1) n 1 1 n 2 is absolutely convergent because 1 n 2 is convergent (p-series with p = 2 > 1). Math114 Lecture 15 7/ 12

15 Example: ( 1) n 1 1 n 2 is absolutely convergent because 1 n 2 is convergent (p-series with p = 2 > 1). Theorem If a series a n is absolutely convergent then it is convergent. Math114 Lecture 15 7/ 12

16 Theorem If a series a n is absolutely convergent then it is convergent. Example Determine whether the following series are absolutely convergent or conditionally convergent. 1 sin(n) 2 n Math114 Lecture 15 8/ 12

17 Theorem If a series a n is absolutely convergent then it is convergent. Example Determine whether the following series are absolutely convergent or conditionally convergent. 1 2 sin(n) 2 n ( 1)n 1 n n 2 +1 Math114 Lecture 15 8/ 12

18 The Ratio Test a (i) If lim n+1 n a n = L < 1, then a n is absolutely convergent and therefore convergent). Math114 Lecture 15 9/ 12

19 The Ratio Test a (i) If lim n+1 n a n = L < 1, then a n is absolutely convergent and therefore convergent). a (ii) If lim n+1 n a n = L > 1 or a n+1 limn a n =, then a n is divergent. Math114 Lecture 15 9/ 12

20 The Ratio Test a (i) If lim n+1 n a n = L < 1, then a n is absolutely convergent and therefore convergent). a (ii) If lim n+1 n a n = L > 1 or a n+1 limn a n =, then a n is divergent. a (iii) If lim n+1 n a n = 1, the Ratio Test is inconclusive. Math114 Lecture 15 9/ 12

21 The Ratio Test a (i) If lim n+1 n a n = L < 1, then a n is absolutely convergent and therefore convergent). a (ii) If lim n+1 n a n = L > 1 or a n+1 limn a n =, then a n is divergent. a (iii) If lim n+1 n a n = 1, the Ratio Test is inconclusive. Idea of proof of item (i): compare series to a convergent geometric series and apply Comparison Test. Idea of proof of item (ii): eventually, > 1, i.e. a n+1 a n eventually a n+1 > a n. So, lim n a n 0. The Divergence Test implies that a n diverges. Math114 Lecture 15 9/ 12

22 Examples Use the ratio test to determine whether the series is convergent or divergent, where possible. 3 n! 100 n Math114 Lecture 15 10/ 12

23 Examples Use the ratio test to determine whether the series is convergent or divergent, where possible. 3 4 n! 100 n 3n 2 n n 3 Math114 Lecture 15 10/ 12

24 Examples Use the ratio test to determine whether the series is convergent or divergent, where possible n! 100 n 3n 2 n n 3 2n 3 n n 3 Math114 Lecture 15 10/ 12

25 Examples Use the ratio test to determine whether the series is convergent or divergent, where possible n! 100 n 3n 2 n n 3 2n 3 n n 3 1 n 3 Math114 Lecture 15 10/ 12

26 The Root Test (i) If lim n n an = L < 1, then a n is absolutely convergent (and therefore convergent). Math114 Lecture 15 11/ 12

27 The Root Test (i) If lim n n an = L < 1, then a n is absolutely convergent (and therefore convergent). (ii) If lim n n an = L > 1 or lim n n an =, then a n is divergent. Math114 Lecture 15 11/ 12

28 The Root Test (i) If lim n n an = L < 1, then a n is absolutely convergent (and therefore convergent). (ii) If lim n n an = L > 1 or lim n n an =, then a n is divergent. (iii) If lim n n an = 1, then the Root Test is inconclusive. Math114 Lecture 15 11/ 12

29 The Root Test (i) If lim n n an = L < 1, then a n is absolutely convergent (and therefore convergent). (ii) If lim n n an = L > 1 or lim n n an =, then a n is divergent. (iii) If lim n n an = 1, then the Root Test is inconclusive. Examples Use the Root Test to determine whether the following series converge or diverge, if possible. 7 ( 2) n n n. Math114 Lecture 15 11/ 12

30 The Root Test (i) If lim n n an = L < 1, then a n is absolutely convergent (and therefore convergent). (ii) If lim n n an = L > 1 or lim n n an =, then a n is divergent. (iii) If lim n n an = 1, then the Root Test is inconclusive. Examples Use the Root Test to determine whether the following series converge or diverge, if possible. 7 8 ( 2) n n. n (arctan(n))n. Math114 Lecture 15 11/ 12

31 Why does rearranging terms in Alternating Harmonic Series yield different sums? Math114 Lecture 15 12/ 12

32 Why does rearranging terms in Alternating Harmonic Series yield different sums? Riemann proved: If a n is conditionally convergent, and r is any real number, then there is a rearrangement of a n whose sum equals r. Math114 Lecture 15 12/ 12

33 Why does rearranging terms in Alternating Harmonic Series yield different sums? Riemann proved: If a n is conditionally convergent, and r is any real number, then there is a rearrangement of a n whose sum equals r. Absolutely convergent series behave more like finite sums! Math114 Lecture 15 12/ 12