SUMMARY OF SEQUENCES AND SERIES

Similar documents
Testing for Convergence

In this section, we show how to use the integral test to decide whether a series

Solutions to Practice Midterms. Practice Midterm 1

Ans: a n = 3 + ( 1) n Determine whether the sequence converges or diverges. If it converges, find the limit.

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

MAT1026 Calculus II Basic Convergence Tests for Series

CHAPTER 10 INFINITE SEQUENCES AND SERIES

Convergence: nth-term Test, Comparing Non-negative Series, Ratio Test

INFINITE SEQUENCES AND SERIES

M17 MAT25-21 HOMEWORK 5 SOLUTIONS

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

Math 113 Exam 3 Practice

= lim. = lim. 3 dx = lim. [1 1 b 3 ]=1. 3. Determine if the following series converge or diverge. Justify your answers completely.

Chapter 6 Infinite Series

Math 113 Exam 4 Practice

BC: Q401.CH9A Convergent and Divergent Series (LESSON 1)

Chapter 10: Power Series

1 Lecture 2: Sequence, Series and power series (8/14/2012)

Series Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1

Math 132, Fall 2009 Exam 2: Solutions

MTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #20 Sakai Web Project Material

Math 106 Fall 2014 Exam 3.2 December 10, 2014

Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan

Definition An infinite sequence of numbers is an ordered set of real numbers.

Infinite Sequence and Series

2 n = n=1 a n is convergent and we let. i=1

Additional Notes on Power Series

JANE PROFESSOR WW Prob Lib1 Summer 2000

Math 106 Fall 2014 Exam 3.1 December 10, 2014

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

Are the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1

The Interval of Convergence for a Power Series Examples

ENGI Series Page 6-01

INFINITE SEQUENCES AND SERIES

MTH 246 TEST 3 April 4, 2014

n=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME

Section 1.4. Power Series

Math 113 Exam 3 Practice

11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series

REVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.

9.3 The INTEGRAL TEST; p-series

6.3 Testing Series With Positive Terms

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

Sequences. A Sequence is a list of numbers written in order.

MATH 312 Midterm I(Spring 2015)

Solutions to Final Exam Review Problems

1 Introduction to Sequences and Series, Part V

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!

Practice Test Problems for Test IV, with Solutions

Power Series: A power series about the center, x = 0, is a function of x of the form

Carleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below.

INFINITE SERIES PROBLEMS-SOLUTIONS. 3 n and 1. converges by the Comparison Test. and. ( 8 ) 2 n. 4 n + 2. n n = 4 lim 1

Ma 530 Introduction to Power Series

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its

Chapter 6: Numerical Series

Math 113 (Calculus 2) Section 12 Exam 4

Notice that this test does not say anything about divergence of an alternating series.

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

AP Calculus Chapter 9: Infinite Series

The Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1

MAS111 Convergence and Continuity

NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed :

11.6 Absolute Convergence and the Ratio and Root Tests

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Section 11.8: Power Series

MIDTERM 2 CALCULUS 2. Monday, October 22, 5:15 PM to 6:45 PM. Name PRACTICE EXAM

Sequences and Series of Functions

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

MTH 133 Solutions to Exam 2 November 16th, Without fully opening the exam, check that you have pages 1 through 12.

MATH 1080: Calculus of One Variable II Fall 2017 Textbook: Single Variable Calculus: Early Transcendentals, 7e, by James Stewart.

Taylor Series (BC Only)

CHAPTER 1 SEQUENCES AND INFINITE SERIES

Math 163 REVIEW EXAM 3: SOLUTIONS

Quiz. Use either the RATIO or ROOT TEST to determine whether the series is convergent or not.


Part I: Covers Sequence through Series Comparison Tests

2.4.2 A Theorem About Absolutely Convergent Series

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:

Chapter 7: Numerical Series

n 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.

5 Sequences and Series

MTH 133 Solutions to Exam 2 Nov. 18th 2015

An alternating series is a series where the signs alternate. Generally (but not always) there is a factor of the form ( 1) n + 1

11.5 Alternating Series; 11.6 Convergence, Ratio, Root Tests

MA131 - Analysis 1. Workbook 9 Series III

Ma 530 Infinite Series I

Solutions to quizzes Math Spring 2007

TRUE/FALSE QUESTIONS FOR SEQUENCES

Please do NOT write in this box. Multiple Choice. Total

Transcription:

SUMMARY OF SEQUENCES AND SERIES Importat Defiitios, Results ad Theorems for Sequeces ad Series Defiitio. A sequece {a } has a limit L ad we write lim a = L if for every ɛ > 0, there is a correspodig iteger N such that if > N the a L < ɛ. Theorem. If lim x f(x) = L ad f() = a, whe is a iteger, the lim a = L. Theorem 2. If lim a = 0, the lim a = 0. Result. The sequece {r } is coverget if < r ad diverget for all other values of r. 0 if < r < lim r = if r = Theorem 3. Every bouded, mootoic sequece is coverget Defiitio. Give a series a = a + a 2 + + a +, let s deote the fiite its th partial sum s = If the sequece {s } is coverget ad lim s = s exists as a real umber, the the series a is called coverget ad we write a a = s The umber s is called the sum of the series. If the sequece {s } is diverget, the the series is called diverget. Result (Geometric Series). The geometric series ar = a + ar + ar 2 +

2 SUMMARY OF SEQUENCES AND SERIES is coverget if r < ad its sum is ar = a r If r, the the geometric series is diverget. r < Theorem 4. If the series a is coverget, the lim a = 0. Theorem 5 (The Test for Divergece). If lim a does ot exist of if lim a 0, the the series a is diverget. Result. If a ad b are coverget series, the so are the series c a, (a + b ) ad (a b ), ( c is a costat) ad () ca = c a (2) (a ± b ) = a ± b Result (The Itegral Test). Suppose f is a cotiuous, positive, decreasig fuctio o [, ) ad let a = f(). The () If f(x) dx is coverget, the a is coverget. (2) If f(x) dx is diverget, the a is diverget. Result (p-series test). The p-series The series is called the Harmoic Series. is coverget if p > ad diverget if p. p Result (The Compariso Test). Suppose that a ad b are series with positive terms. () If b is coverget ad a b for all, the a is also coverget. (2) If b is diverget ad a b for all, the a is also diverget. Result (The Limit Compariso Test). Suppose that a ad b are series with positive terms. If a lim = c b where c is a fiite umber ad c > 0, the either both series coverge or both diverge.

SUMMARY OF SEQUENCES AND SERIES 3 Defiitio. A alteratig series is a series whose terms are alteatively positive ad egative. Result (The Alteratig Series Test). A alteratig series ( ) b = b b 2 + b 3 + b > 0 is coverget if it satisfies the followig coditios () b + b for all (2) lim b = 0. Defiitio (Absolutely Covergece). A series a is called absolutely coverget if the series of the absolute values a is coverget. Defiitio. A series a is called coditioally coverget if it is coverget but ot absolutely coverget. Theorem 6. If a series a is absolutely coverget, the it is coverget. Result (The Ratio Test). Give a series a, () If lim a + a = L <, the the series a is absolutely coverget. (2) If lim a + a = L > or lim a + a =, the the seriest a is diverget. (3) If lim a + a = L =, the ratio Test is icoclusive. Result (The Root Test). Give a series a, () If lim a = L <, the the series a is absolutely coverget. (2) If lim a = L > or lim (3) If lim a = L =, the root Test is icoclusive. a =, the the seriest a is diverget. Defiitio (Power Series). A power series, cetered at a, is of the form c (x a) = c 0 + c (x a) + c 2 (x a) 2 +, where a is a fixed umber, x is a variable. c s are costats, called the coefficiets of the series. Theorem 0.. For a give series c (x a), there are oly three posssibilities:

4 SUMMARY OF SEQUENCES AND SERIES () The series oly coverges for x = a. (2) The series coverges for all values of x. (3) There is a positive umber R such that the seies coverges whe x a < R ad diverges if x a > R. The umber R i (3) is called the radius of covergece of the power series. I (), R = 0, while i (2), R =. Remark. Note that the series may or may ot coverge if x a = R. What happes at these poits will ot chage the radius of covergece. Defiitio. The iterval of all x s, icludig the edpoits if eed be, for which the power series coverges is called the iterval of covergece of the series. For (), the iterval of covergece is {a}. For (2), the iterval of covergece is (, ). For (3), the iterval of covergece is either ( a R, a+r ), ( a R, a+r ], [ a R, a+r ) or [ a R, a+r ], depedig whether the series coverges for the edpoits of the iterval. Theorem 0.2. If f has a power series represetatio at a, that is, if f(x) = c (x a) x a < R the it s coefficiets are give by the formula c = f () (a) I other words, if f has a power series expasio at a, the it must be of the form f(x) = f () (a) (x a) This series is called the Taylor series of fuctio about a (or cetered at a). Defiitio (Maclauri Series). I the special case that a = 0, the series f(x) = c x where c = f () (0), is called the Maclauri series of the fuctio f(x). Importat Maclauri Series The followig Maclauri series are very importat ad you are expected to remember them: () x = + x + x2 + x 3 + = x, < x <. (2) + x = x + x2 x 3 + = ( ) x, < x <.

(3) e x = + x + x2 2! + x3 3! + = SUMMARY OF SEQUENCES AND SERIES 5 (4) cos(x) = x2 2! + x4 4! x6 6! + = x, x i (, ). (5) si(x) = x x3 3! + x5 5! x7 7! + = ( ) x 2 (2)! ( ) x 2+ (2 + )!, x i (, )., x i (, ) Remark. To avoid cofusio betwee the series for si x ad cos x, remember that: si x is a odd fuctio, that is, si( x) = si x, so the series for si x has oly odd expoets. cos x is a eve fuctio, that is cos( x) = cos x, so the series for cos x has oly eve expoets.