( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

Similar documents
Math 132, Fall 2009 Exam 2: Solutions

1 Introduction to Sequences and Series, Part V

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

6.3 Testing Series With Positive Terms

10.6 ALTERNATING SERIES

E. Incorrect! Plug n = 1, 2, 3, & 4 into the general term formula. n =, then the first four terms are found by

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

Infinite Sequences and Series

Review for Test 3 Math 1552, Integral Calculus Sections 8.8,

CALCULUS II. Sequences and Series. Paul Dawkins

9.3 Power Series: Taylor & Maclaurin Series

Midterm Exam #2. Please staple this cover and honor pledge atop your solutions.

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.

10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.

Math 113 Exam 3 Practice

Section 11.8: Power Series

Math 25 Solutions to practice problems

10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

Sequences, Series, and All That

Solutions to Homework 7

MA131 - Analysis 1. Workbook 7 Series I

CALCULUS II Sequences and Series. Paul Dawkins

Math 113 Exam 3 Practice

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.

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

Math 1314 Lesson 16 Area and Riemann Sums and Lesson 17 Riemann Sums Using GeoGebra; Definite Integrals

Complex Numbers Primer

Roberto s Notes on Series Chapter 2: Convergence tests Section 7. Alternating series

Complex Numbers Primer

Ma 530 Introduction to Power Series

Practice Test Problems for Test IV, with Solutions

Testing for Convergence

Math 116 Practice for Exam 3

CHAPTER 10 INFINITE SEQUENCES AND SERIES

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

5 Sequences and Series

11.6 Absolute Convergence and the Ratio and Root Tests

Spring 2016 Exam 2 NAME: PIN:

INFINITE SEQUENCES AND SERIES

Math 31B Integration and Infinite Series. Practice Final

Lecture 2 Appendix B: Some sample problems from Boas, Chapter 1. Solution: We want to use the general expression for the form of a geometric series

MTH 133 Solutions to Exam 2 Nov. 18th 2015

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

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

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

MA131 - Analysis 1. Workbook 10 Series IV

Chapter 6: Numerical Series

Lecture 3: Catalan Numbers

Math 113 Exam 4 Practice

Chapter 7: Numerical Series

Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer)

f x x c x c x c... x c...

MATH 2300 review problems for Exam 2

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

Infinite Sequence and Series

MA131 - Analysis 1. Workbook 9 Series III

10.2 Infinite Series Contemporary Calculus 1

Series III. Chapter Alternating Series

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

Example 2. Find the upper bound for the remainder for the approximation from Example 1.

Calculus II Homework: The Comparison Tests Page 1. a n. 1 n 2 + n + 1. n= n. n=1

MAT1026 Calculus II Basic Convergence Tests for Series

MATH 31B: MIDTERM 2 REVIEW

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

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

Section 5.5. Infinite Series: The Ratio Test

Fall 2018 Exam 2 PIN: 17 INSTRUCTIONS

Sequences. Notation. Convergence of a Sequence

Riemann Sums y = f (x)

Part I: Covers Sequence through Series Comparison Tests

Series: Infinite Sums

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)

If we want to add up the area of four rectangles, we could find the area of each rectangle and then write this sum symbolically as:

MATH 312 Midterm I(Spring 2015)

5.6 Absolute Convergence and The Ratio and Root Tests

Ma 530 Infinite Series I

Math 106 Fall 2014 Exam 3.2 December 10, 2014

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

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

1 Generating functions for balls in boxes

Math 10A final exam, December 16, 2016

MATH2007* Partial Answers to Review Exercises Fall 2004

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

Section 7 Fundamentals of Sequences and Series

0.1. Geometric Series Formula. This is in your book, but I thought it might be helpful to include here. If you have a geometric series

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

Math 113, Calculus II Winter 2007 Final Exam Solutions

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

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

Section 1.4. Power Series

Math 116 Practice for Exam 3

Math 312 Lecture Notes One Dimensional Maps

1. Do the following sequences converge or diverge? If convergent, give the limit. Explicitly show your reasoning. 2n + 1 n ( 1) n+1.

SYDE 112, LECTURE 2: Riemann Sums

Infinite Series and Improper Integrals

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

Fooling Newton s Method

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

MATH 1910 Workshop Solution

Transcription:

Preface Here are the solutios to the practice problems for my otes. Some solutios will have more or less detail tha other solutios. As the difficulty level of the problems icreases less detail will go ito the basics of the solutio uder the assumptio that if you ve reached the level of workig the harder problems the you will probably already uderstad the basics fairly well ad wo t eed all the explaatio. This documet was writte with presetatio o the web i mid. O the web most solutios are broke dow ito steps ad may of the steps have hits. Each hit o the web is give as a popup however i this documet they are listed prior to each step. Also, o the web each step ca be viewed idividually by clickig o liks while i this documet they are all showig. Also, there are liable to be some formattig parts i this documet iteded for help i geeratig the web pages that have t bee removed here. These issues may make the solutios a little difficult to follow at times, but they should still be readable. Sequeces. List the first 5 terms of the followig sequece. 7 = 0 Solutio There really is t all that much to this problem. All we eed to do is, startig at = 0, plug i the first five values of ito the formula for the sequece terms. Doig that gives, 0 = 0: = 0 0 7 = : 7 6 8 8 = : 7 = : = = 6 7 6 = : = 7 9 So, the first five terms of the sequece are, 007 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx

8 6 0,,, 6,, 9 Note that we put the formal aswer iside the braces to make sure that we do t forget that we are dealig with a sequece ad we made sure ad icluded the at the ed to remider ourselves that there are more terms to this sequece that just the five that we listed out here.. List the first 5 terms of the followig sequece. + ( ) + = Solutio There really is t all that much to this problem. All we eed to do is, startig at =, plug i the first five values of ito the formula for the sequece terms. Doig that gives, So, the first five terms of the sequece are, + + + + + + 5+ + 6+ + = = : = : 89 89 = 5: 5 5 = 6: 6 6 7 7,,,,, 89 7 Note that we put the formal aswer iside the braces to make sure that we do t forget that we are dealig with a sequece ad we made sure ad icluded the at the ed to remider ourselves that there are more terms to this sequece that just the five that we listed out here. 007 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx

. Determie if the give sequece coverges or diverges. If it coverges what is its limit? 7+ + 0 = Step To aswer this all we eed is the followig limit of the sequece terms. 7+ lim = 0 + You do recall how to take limits at ifiity right? If ot you should go back ito the Calculus I material do some refreshig o limits at ifiity as well at L Hospital s rule. Step We ca see that the limit of the terms existed ad was a fiite umber ad so we kow that the sequece coverges ad its limit is.. Determie if the give sequece coverges or diverges. If it coverges what is its limit? ( ) + Step To aswer this all we eed is the followig limit of the sequece terms. lim ( ) = 0 + However, because of the we ca t compute this limit usig our kowledge of computig limits from Calculus I. Step Recall however, that we had a ice Fact i the otes from this sectio that had us computig ot the limit above but istead computig the limit of the absolute value of the sequece terms. ( ) lim = lim = 0 + + This is a limit that we ca compute because the absolute value got rid of the alteratig sig, i.e. the +. 007 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx

Step Now, by the Fact from class we kow that because the limit of the absolute value of the sequece terms was zero (ad recall that to use that fact the limit MUST be zero!) we also kow the followig limit. ( ) lim = 0 + Step We ca see that the limit of the terms existed ad was a fiite umber ad so we kow that the sequece coverges ad its limit is zero. 5. Determie if the give sequece coverges or diverges. If it coverges what is its limit? e 5 e = Step To aswer this all we eed is the followig limit of the sequece terms. 5 5 e 5 5 lim lim lim = e = e e e = You do recall how to use L Hospital s rule to compute limits at ifiity right? If ot you should go back ito the Calculus I material do some refreshig. Step We ca see that the limit of the terms existed ad but was ifiite ad so we kow that the sequece diverges. 6. Determie if the give sequece coverges or diverges. If it coverges what is its limit? ( + ) ( + ) l l Step To aswer this all we eed is the followig limit of the sequece terms. = l ( + ) lim lim + = + = lim = l ( + ) ( + ) + 007 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx

You do recall how to use L Hospital s rule to compute limits at ifiity right? If ot you should go back ito the Calculus I material do some refreshig. Step We ca see that the limit of the terms existed ad was a fiite umber ad so we kow that the sequece coverges ad its limit is oe. 007 Paul Dawkis 5 http://tutorial.math.lamar.edu/terms.aspx