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Advanced Placement Infinite Chapter 10 Section 5 Testing Convergence at Endpoints

Essential Question: How can the convergence or of a series be determined on an interval including the endpoints?

Student Objectives: The student will be able to use the nth- and the to determine convergence or divergence of a series of numbers.

The student will be able to determine convergence or divergence of p- series, including the harmonic series.

The student will be able to determine the convergence, , or divergence of a at the endpoints of the interval of convergence.

Terms: Alternating series Comparison Test Conditional Convergence Harmonic Series Test Intervals of Convergence p-series

Key Concepts: A word of Caution Absolute and Conditional Convergence Alternating Series Comparison Test Harmonic Series and p-Series Integral Test Intervals of Convergence

Theorems: Theorem 10: The Interval Test

Let a be a of positive terms. Suppose a f n , where f is a continuous, { n } n = ( ) positive, decreasing function of x for all x ! N !(N is a positive integer). Then the

" " series an and the integral f (x)dx!either both converge or both diverge.! # $N n= N

Theorem 11:!!The (LCT)

Suppose that an > 0!and bn > 0!for all n"N (N is a positive integer).

an 1. If! lim = c,!0 < c < ",!then! an !and! bn !both converge or diverge. x!" # # bn

an 2. If! lim = 0!and! bn converges,!then an !converges. !! x!" b # # n an 3. If! lim = "!and! bn diverges,!then an !diverges.!!!!!!!!!!!!!!!!!!!!!!!!!!! x!" # # bn

Theorem 12:!!The Alternating Series!Test!(Leibnitz's Theorem) The series " n+1 #(!1) un = u1 ! u2 + u3 ! u4 + u5 ! u6 +… n=1 converges if all three of the following conditions are satisfied:

1. each un !is positive;!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 2. u $ u !for all n $ N,!for some integer N; n n+1 3. lim un % 0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! n%"

Theorem 13:!!The Alternating Series Estimation Theorem ! n+1 If the alternating series "(-1) un , satisfies the conditions of Theorem 12, then the n=1

truncation error for the nth partial sum is less than un and has the same sign as the unused term.

Definitions:

Rearrangements of Absolutely

If !!an !converges absolutely, and if b1,!b2 ,!b3,!b4 ,!…!bn ,!…! is an rearrangement " "

of the series {an }, then !bn converges absolutely and !bn = !an . n=1 n=1

Rearrangements of Conditionally Convergent Series

If !!an !converges conditionally, then the terms can be rearranged to form a

. The terms can also be rearranged to form a series that converges

to any preassigned sum.

Procedures

" n How to Test a Power Series #cn (x ! a) for Convergence n=0 1.!!Use the to find the values of x for which the series converges !!!!!absolutely. Ordinarily, this is an open interval:!!a ! R < x < a + R,!In some !!!!!instantances, the series converges for all values of x. In rare cases, the series

!!!!!converges only at x = a.

2. If the interval for absolute convergence is finite, test for convergence or

!!!!!divergence at each endpoint. The Ratio Test fails at these points. Use a

!!!!!Comparison Test, the Integral Test, or the Alternating Series Test.

3. If the interval of absolute convergence is a ! R < x < a + R, conclude that the

!!!!!series diverges (it does not even converge consitionally) for x ! a > R, because for !!!!!those values fo x the nth term does not apporach zero.

Procedures for Determining Convergence

Geometric Series Test

nth Term Test Is! a = ! n YES 1 YES Conv erges!to! !if ! r < 1 Is!lim an = 0? 2 3 1! r n!" a + ar + ar + ar Diverges!if ! r " 1 +…?

NO NO

Series!diverges Does the series YES Series!converges!if ! p > 1 have the form Series!diverges!if ! p ! 1 ! 1 p-Series Test ? " n p n=1

NO

Apply one of the comparison tests,

Integral Test, Ratio Test, Does an ! YES ! O riginal!Series!converges

nth . Test converge? nonnegative terms and/or absolute

convergence. NO

Is!N !!! Alternating Is! a YES ! n such!that YES Series Test alternating? u u ...? N " N =1 "

NO NO

See what you can do with the partial sums, consult more advanced books, or explore with a CAS.

Series!converges!if !u 0 Graphing Calculator Skills: n !

Series!diverges!if !un ! 0 None

Sample Questions: ! 1 " p 1. Use the Integral Test to prove that n=1 n converges for p > 1.

! 1 " p 2. Use the Integral Test to prove that n=1 n converges for p < 1.

! 1 " p 3. Use the Integral Test to prove that n=1 n converges for p = 1.

# 0 x = 0 % 1 Let f (x) = $ ! x2 &%e x " 0 It can be shown (although not easily) that f has of all orders at (n) x = 0 and that f (0) = 0 for all n. Use this fact as you proceed with the next three questions.

4. Construct the Maclaurin series for the f .

5. For what values of x does the series converge?

6. Find all the values of x for which the series actually converges to f (x).

7. Determine whether the converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! 1 dx " 4 1 x 3

8. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! x2 dx " x3 + 1 1

9. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! ln x dx " x 1

10. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! 1+ cos x dx " x2 1

11. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! x dx " 1+ x 1

12. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 3 f (x) = x

13. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 7x f (x) = 2 x ! 8

14. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 3 + x2 f (x) = 2 3 ! x

15. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. sin x f (x) = 5 x

16. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. ! 1$ f (x) = ln# & " x%

Homework: Pages 528 - 530

Exercises: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, and 69 Exercises: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, and 70 Exercises: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, and 71 Exercises: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, and 68

Sample Questions: ! 1 " p 1. Use the Integral Test to prove that n=1 n converges for p > 1. 1 Note that the Integral Test applies to any series of the form p where p is positive. ! n - p This is because the function f (x) = x , is continuous and positive for x > 0, and f '(x) = " px" p"1 is negative for all x > 0.

# 1 If ! p > 1: dx $ p 1 x b 1 lim dx b%# $ x p 1 b , & 1 ) / lim . ( p"1 + 1 b%# . " p + 1 x 1 - '( ) * 1 0

, / & 1 ) & 1 ) lim . ( p"1 + " ( p"1 +1 b%# - '((" p + 1)(b) *+ '((" p + 1)(1) *+0

, & 1 ) & 1 )/ lim . ( p"1 + " ( +1 b%# - '((" p + 1)(b) *+ '(" p + 1)(1)*0

, & 1 ) & 1 )/ lim ( p 1 + " b . " ( +1 %# - ((" p + 1)(b) + '(" p + 1)*0 ' * , 1 1 / lim . p"1 " 1 b%# - (" p + 1)(b) (" p + 1)0

1 1 b 1 " (" p + 1)(#) " (" p + 1) 1 1 " (" p + 1)(#) (" p + 1) 1 1 " # (" p + 1)

1 0 " (" p + 1) "1 " p + 1 1 1 !Since! < #,!the!series! p " 1 p " 1 converges by the Integral Test.

! 1 " p 2. Use the Integral Test to prove that n=1 n converges for p < 1.

! 1 If !0 < p < 1: dx " p 1 x 1

p 1 #

1 $Since! % !,!The series p # 1

diverges by the Integral Test. If p & 0, the series diverges by the nth Term Test. This now completes the proof for p < 1.

! 1 " p 3. Use the Integral Test to prove that n=1 n converges for p = 1. ! 1 If ! p = 1: " dx 1 x b 1 lim dx b#! " 1 x b lim [ln x] b#! ( 1 ) lim lnb $ ln x b#! ([ ] [ ]) lim lnb $ 0 b#! ([ ] [ ]) lim(lnb) b#! ln(!) ! ! 1 %Since!" dx # !,!The series 1 x diverges by the Integral Test.

# 0 x = 0 % 1 Let f (x) = $ ! x2 &%e x " 0 It can be shown (although not easily) that f has derivatives of all orders at (n) x = 0 and that f (0) = 0 for all n. Use this fact as you proceed with the next three questions.

4. Construct the Maclaurin series for the f .

Since f (n) (0) = 0 for all n, the Maclaurin series for f has all zero coefficients! ! The series is simply "(0 ! xn ) = 0. n=0

5. For what values of x does the series converge?

The series coverges for all values of x. The series converges to 0.

6. Find all the values of x for which the series actually converges to f (x).

Since f (x) = 0 only at x = 0, the only place that this series actually converges to its f -value is at x = 0.

7. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! 1 dx " 4 1 x 3 ! 1 Since the integral is in the form dx!of " p 1 x and p < 1, the series converges.

8. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! x2 dx " x3 + 1 1 1 Use!the!Limit!Comparison!Test!with! . x x2 3 lim x + 1 x!" 1 x $ x2 x' lim # x!" %& x3 + 1 1() $ x3 ' lim x!" %& x3 + 1() $ 3x2 ' lim x!" %& 3x2 () lim(1) x!" 1 *Since the Limit Comparison Test results in a limit of 1, the must " 1 " x2 both converge or diverge. Since dx!diverges, dx!must also diverge. + x + x3 + 1 1 1

9. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! ln x dx " x 1 1 Use!the!Limit!Comparison!Test!with! . x ln x lim x x!" 1 x $ ln x x' lim & # ) x!" % x 1( lim(ln x) x!" ln(") " *Since the Limit Comparison Test results in a limit of " " 1 " x2 and dx!diverges,!then dx!must also diverge. + x + x3 + 1 1 1

10. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! 1+ cos x dx " x2 1 2 Use!the!Limit!Comparison!Test!with! . x2 1+ cos x 2 lim x x!" 2 x2 $ 1+ cos x x2 ' lim # x!" %& x2 2 () $ 1+ cos x' lim & ) x!" % 2 ( 1+ cos(") 2 1+ (±1) 2 Oscillates!between 1* 1 1+ 1 2 2 0 2 2 2 0 1

+Since the Limit Comparison Test results in a limit between 0 and 1 " 2 " 1+ cos x and dx!converges,!then dx!must also converge. , x2 , x2 1 1

11. Determine whether the improper integral converges or diverges. Give reasons for your answer. DO NOT evaluate the integral. ! x dx " 1 x 1 +

1 Use!the!Limit!Comparison!Test!with! . x x lim 1+ x x!" 1 x $ x x ' lim & # ) x!" % 1+ x 1 ( $ x ' lim & ) x!" % 1+ x( $ 1' lim & ) x!" % 1( lim(1) x!" 1

*Since the Limit Comparison Test results in a limit of 1 both integrals " 1 " x must converge or diverge. Since dx!diverges,!then dx!must also diverge. + x + 1+ x 1 1

12. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 3 f (x) = x Yes,!!N = 0.

13. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 7x f (x) = 2 x ! 8

Yes,!N = 2 2 !or!2.8283. So,!let!N = 3.

14. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. 3 + x2 f (x) = 2 3 ! x

No,!the function is negative and increasing for x > 3 = 1.7321.

15. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. sin x f (x) = 5 x

No,!the function oscilllates.

16. Determine whether the function is positive, decreasing on some interval (N,!!). You do not need to identify the value of N. ! 1$ f (x) = ln# & " x%

No,!the function is not positive for x ! 1.