Math Absolute Convergence, Ratio Test, Root Test

Transcription

1 Math Absolute Convergence, Ratio Test, Root Test Peter A. Perry University of Kentucky February 20, 2017

2 Bill of Fare 1. Review and Recap 2. Dirichlet s Dilemma 3. Absolute Convergence 4. Ratio Test 5. Root Test

3 Sequences and Series A sequence is an infinite list of numbers a 1, a 2,..., a n,..., one for each positive integer A series i=1 a i represents the sum of an infinite list of numbers a 1, a 2, a 3,..., a n,... The nth partial sum of an infinite series i=1 a i is the finite sum s n = n i=1 a i = a 1 + a a n The series i=1 a i converges if the sequence of partial sums has a limit as n. s 1, s 2, s 3,..., s n,...

4 Converge or Diverge? A certain series n=1 a n has the property that lim n a n = 1. Does this series converge or diverge? A. Converge B. Diverge

5 It s the Tail That Matters A simple but important fact about series is: If n=100 a n converges, then n=1 a n also converges.

6 It s the Tail That Matters A simple but important fact about series is: If n=100 a n converges, then n=1 a n also converges. And there is nothing special about 100. In fact: If M is a positive integer and n=m+1 a n converges, then n=1 a n also converges.

7 Absolute Convergence A series n=1 a n is called absolutely convergent if n=1 a n converges. Example 1: The series 1 n=1 ( 1)n 1 is convergent, but not n absolutely convergent. Such a series is called conditionally convergent. Example 2: The series 1 n=1 ( 1)n 1 is both convergent and n2 absolutely convergent. Roughly speaking a series is absolutely convergent if the convergence isn t due to cancellations between positive and negative terms.

8 The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever. By using them, one may draw any conclusion he pleases and that is why these series have produced so many fallacies and so many paradoxes. Neils Hendrik Abel, 1826 The theory of infinite series was developed in the nineteenth century. Let s look at one of the paradoxes that led to the notion of absolute convergence that we ll study today.

9 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

10 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

11 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = S = Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

12 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = S = Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

13 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = S = S = Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

14 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = S = S = S = S Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

15 Dirichlet s Dilemma In 1827, Pierre Lejeune Dirichlet discovered the following paradox while studying infinite series: S = S = S = Thus rearrangement of a conditionally convergent series leads to fallacies and paradoxes! Source: Mathematics Teacher, November 1987, Volume 80, Number 8, pp

16 Absolute Convergence A series n=1 a n is called absolutely convergent if n=1 a n converges. Example 1: The series 1 n=1 ( 1)n 1 is convergent, but not n absolutely convergent. Such a series is called conditionally convergent. Example 2: The series 1 n=1 ( 1)n 1 is both convergent and n2 absolutely convergent. Roughly speaking a series is absolutely convergent if the convergence isn t due to cancellations between positive and negative terms.

17 Dirichlet s Dilemma and Riemann s Resolution Bernhard Riemann ( ) proved that a conditionally convergent series can be rearranged to sum to any real number! On the other hand, it can be shown that any rearrangment of an absolutely convergent series gives the same sum

18 Converge Conditionally, Converge Absolutely, or Diverge? Which of the following statements is true about the series n=1 ( 1)n 1 1 n 1/2? A. The series converges absolutely B. The series converges conditionally. C. The series diverges.

19 Converge Conditionally, Converge Absolutely, or Diverge? Which of the following statements is true about the series n=1 ( 1)n 1 1 n 3/2? A. The series converges absolutely B. The series converges conditionally. C. The series diverges.

20 Absolute Convergence Theorem If a series a n is absolutely convergent, it is also convergent. Proof.

21 Absolute Convergence Theorem If a series a n is absolutely convergent, it is also convergent. Proof. First, note that 0 a n + a n 2 a n Since n=1 a n converges, so does n=1 2 a n.

22 Absolute Convergence Theorem If a series a n is absolutely convergent, it is also convergent. Proof. First, note that 0 a n + a n 2 a n Since n=1 a n converges, so does n=1 2 a n. comparison test, n=1 a n + a n converges. Second, by the

23 Absolute Convergence Theorem If a series a n is absolutely convergent, it is also convergent. Proof. First, note that 0 a n + a n 2 a n Since n=1 a n converges, so does n=1 2 a n. comparison test, n=1 a n + a n converges. a n = n=1 n=1 we see that n=1 a n converges. (a n + a n ) n=1 Since a n Second, by the

24 The Ratio Test Remember that the gold standard series n=1 ar n 1 { converges, r < 1 diverges, r 1 The ratio test for convergence of a series n=1 a n focusses on the number L = lim a n+1 n a n If L < 1, then the series n=1 a n behaves like a geometric series with ratio less than one in absolute value (converges) If L > 1 then the series n=1 a n behaves like a geometric series with ratio larger than one in absolute value (diverges)

25 The Ratio Test The Ratio Test a n+1 (i) If lim n a n = L < 1, then n=1 a n converges absolutely. a n+1 (ii) If lim n a n = L > 1, then n=1 a n diverges. a n+1 (iii) If lim n a n = 1, the ratio test is inconclusive Why does it work?

26 Why the Ratio Test Works - Convergence If lim n a n+1 a n so for n N = L < 1, then, if N is large enough a n+1 a n r < 1 for any n N a n a N r n N and n=n+1 a n converges by comparison with a geometric series.

27 Why the Ratio Test Works - Divergence If lim n a n+1 a n so for n N = L > 1, then, if N is large enough a n+1 a n r > 1 for any n N a n a N r n N and lim n a n = so the series can t converge.

28 The Ratio Test The Ratio Test a n+1 (i) If lim n a n = L < 1, then n=1 a n converges absolutely. a n+1 (ii) If lim n a n = L > 1, then n=1 a n diverges. a n+1 (iii) If lim n a n = 1, the ratio test is inconclusive What happens when you try the ratio test on the p series 1 n=1 n p?

29 a n+1 Let L = lim n a n. If L < 1 n=1 a n converges absolutely L > 1 n=1 a n diverges Does the series converge or diverge? n=1 n 5 n

30 a n+1 Let L = lim n a n. If L < 1 n=1 a n converges absolutely L > 1 n=1 a n diverges Does the series converge or diverge? n=1 nπ n ( 3) n 1

31 a n+1 Let L = lim n a n. If L < 1 n=1 a n converges absolutely L > 1 n=1 a n diverges Does the series converge or diverge? cos(nπ/3) n=1 n!

32 The Root Test Another way of looking at our gold standard geometric series is that the terms a 1 = a, a 2 = ar, a 3 = ar 2 behave roughly like a n r n 1. If you take nth roots (a n ) 1/n = a 1/n ( r n 1) 1/n what limit do you get as n? In the root test for convergence of n=1 a n, we consider the limit ( L = lim a n 1/n) n if it exists.

33 The Root Test Another way of looking at our gold standard geometric series is that the terms a 1 = a, a 2 = ar, a 3 = ar 2 behave roughly like a n r n 1. If you take nth roots (a n ) 1/n = a 1/n ( r n 1) 1/n what limit do you get as n? lim n (a n ) 1/n = r In the root test for convergence of n=1 a n, we consider the limit ( L = lim a n 1/n) n if it exists.

34 The Root Test (i) If lim n a n 1/n = L < 1, then the series a n is absolutely convergent (ii) If lim n a n 1/n = L > 1, then the series a n is divergent (iii) If lim n a n 1/n = 1, the root test is inconclusive Legal Notice: a n 1/n and n a n are two names for the same thing

35 The Root Test (i) If lim n a n 1/n = L < 1, then the series a n is absolutely convergent (ii) If lim n a n 1/n = L > 1, then the series a n is divergent (iii) If lim n a n 1/n = 1, the root test is inconclusive What happens when you apply the root test to a p-series?

36 The Root Test (i) If lim n a n 1/n = L < 1, then the series a n is absolutely convergent (ii) If lim n a n 1/n = L > 1, then the series a n is divergent (iii) If lim n a n 1/n = 1, the root test is inconclusive Does the series ( 2) n n=1 n n converge or diverge?

37 The Root Test (i) If lim n a n 1/n = L < 1, then the series a n is absolutely convergent (ii) If lim n a n 1/n = L > 1, then the series a n is divergent (iii) If lim n a n 1/n = 1, the root test is inconclusive Does the series ( 2) n n=1 n n converge or diverge?

38 The Root Test (i) If lim n a n 1/n = L < 1, then the series a n is absolutely convergent (ii) If lim n a n 1/n = L > 1, then the series a n is divergent (iii) If lim n a n 1/n = 1, the root test is inconclusive Does the series n=1 ( n 2 ) n + 1 2n 2 converge or diverge? + 1