MATH 166 TEST 3 REVIEW SHEET

Similar documents
Part I: Covers Sequence through Series Comparison Tests

Math 113 Exam 3 Practice

Math 106 Fall 2014 Exam 3.2 December 10, 2014

Math 132, Fall 2009 Exam 2: Solutions

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

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

Chapter 7: Numerical Series

Chapter 6: Numerical Series

Math 163 REVIEW EXAM 3: SOLUTIONS

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

Testing for Convergence

6.3 Testing Series With Positive Terms

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.

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

MATH 2300 review problems for Exam 2

9.3 The INTEGRAL TEST; p-series

Math 106 Fall 2014 Exam 3.1 December 10, 2014

Chapter 10: Power Series

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

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

MTH 246 TEST 3 April 4, 2014

Math 116 Practice for Exam 3

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

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)

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

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

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

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

Calculus with Analytic Geometry 2

Ma 530 Infinite Series I

Series: Infinite Sums

Math 113 Exam 4 Practice

MAT1026 Calculus II Basic Convergence Tests for Series

Practice Test Problems for Test IV, with Solutions

Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ.

Math 113 (Calculus 2) Section 12 Exam 4

INFINITE SEQUENCES AND SERIES

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

Math 142, Final Exam. 5/2/11.

CHAPTER 10 INFINITE SEQUENCES AND SERIES

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.

Fall 2018 Exam 2 PIN: 17 INSTRUCTIONS

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

Chapter 6 Infinite Series

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

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.

Infinite Sequence and Series

Calculus BC and BCD Drill on Sequences and Series!!! By Susan E. Cantey Walnut Hills H.S. 2006

Strategy for Testing Series

Series III. Chapter Alternating 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.

P-SERIES AND INTEGRAL TEST

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

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

7 Sequences of real numbers

Math 113 Exam 3 Practice

Infinite Sequences and Series

Math 5C Discussion Problems 2 Selected Solutions

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

MATH 21 SECTION NOTES

JANE PROFESSOR WW Prob Lib1 Summer 2000

LECTURES 5 AND 6: CONVERGENCE TESTS

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

MATH 2300 review problems for Exam 2

Math 31B Integration and Infinite Series. Practice Final

MATH 2300 review problems for Exam 2

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and

11.6 Absolute Convergence and the Ratio and Root Tests

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

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

Solutions to Homework 7

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

Series Solutions (BC only)

SUMMARY OF SEQUENCES AND SERIES

Not for reproduction

MA131 - Analysis 1. Workbook 9 Series III

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

ENGI Series Page 6-01

MATH 31B: MIDTERM 2 REVIEW

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 19B Final. Study Aid. June 6, 2011

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Ma 530 Introduction to Power Series

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

Once we have a sequence of numbers, the next thing to do is to sum them up. Given a sequence (a n ) n=1

Section 11.8: Power Series

Infinite Series. Definition. An infinite series is an expression of the form. Where the numbers u k are called the terms of the series.

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

1. C only. 3. none of them. 4. B only. 5. B and C. 6. all of them. 7. A and C. 8. A and B correct

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

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 +

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

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

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

The Interval of Convergence for a Power Series Examples

AP Calculus Chapter 9: Infinite Series

Math 152 Exam 3, Fall 2005

MATH2007* Partial Answers to Review Exercises Fall 2004

Transcription:

MATH 66 TEST REVIEW SHEET Geeral Commets ad Advice: The studet should regard this review sheet oly as a sample of potetial test problems, ad ot a ed-all-be-all guide to its cotet. Aythig ad everythig which we have discussed i class is fair game for the test. The test will cover roughly Sectios 0., 0., 0., 0.4, ad 0.. Do t limit your studyig to this sheet; if you feel you do t fully uderstad a particular topic, the try to do more problems out of your textboo! Summary of Relative Growth Rates. For two sequeces a ) ad b ) such that a ) ad b ), let the symbols a ) << b ) deote that b ) grows faster tha a ). The the followig is a summary of our most commoly ecoutered growth rates: l ) << l ) r ) << q ) << ) << p ) << b ) <<!) << ), where r >, 0 < q <, p >, ad b >. Geeral Strategy for Determiig Covergece of a Series. There is o hard ad fast rule for determiig whether a series coverges or diverges, ad i geeral the studet should build up their ow ituitio o the subject through practice. However, the followig metal checlist ca be a helpful guidelie. ) Divergece Test. Give a series a, the first thig oe should always as is, do the terms a ) 0? If ot, we are doe ad the series diverges by the DT. If the terms do shri to zero, however, we have to put more thought ito the problem. ) Very Nice Series. Is the series a p-series with p >, or a geometric series, or a telescopig series? If so we are doe ad the series coverges by our geeral theorems. Note that sometimes a telescopig series ca be hidde : for example is a telescopig series i disguise. The + ) method of partial fractios may reveal these.) ) Alteratig Series Test. If the series is alteratig ad the terms go to zero, the the series coverges by the AST. 4) Itegral Test. Do the terms of the series loo lie a fuctio we could probably itegrate? As a example, the series + shouts Substitutio Rule whe we cosider its improper itegral aalogue ) x + x ) dx. Remember the series ad the improper itegral are ot equal to oe aother, but they do either both coverge or both diverge.) ) Direct Compariso Test ad Limit Compariso Test. Do the terms a of the series behave similarly to those of oe of the Very Nice Series? If

8 so, a compariso test may be appropriate. Usually the LCT is more powerful tha the DCT. For examples, the series + seems to 7 behave lie for very large ad so we thi it probably coverges; 7 + 00 coversely the series seems to behave lie a costat multiple of the harmoic series for very large, so we thi it diverges. Usually the atural choices for a compariso test are a p-series or a geometric series.) 6) Ratio Test. Are we looig at a series of fractios where the questio of covergece seems to come dow to comparig the growth rate of the umerator vs. the deomiator? e.g. or l ) 4 or 000.! The Ratio Test has a good chace of worig. Ratio Test is almost always the most atural choice if there is a factorial! term appearig aywhere i the series.) 7) Root Test. Are we looig at a series where to the -th power occurs ) i a sigificat way? e.g. or + ). Root Test + =0 might be easiest. 8) Words of Warig. Your first attempt at a test may ot be the right choice. Be aware of whe a test fails for istace if the limit i Ratio/Root Test is exactly, or extremal cases of Limit Compariso Test where limit is 0 or, etc.) ad be prepared to try a differet strategy. O that ote, it is ofte very helpful to do some private scratchwor where you determie for yourself if a particular test solves the problem, ad the carefully write a solutio for full credit i a homewor, quiz, or exam settig.. Determie if the sequece coverges, ad if so, state to what limit. a) 4 b) arcta000) c) e 000 = d) + e) f) arcta 7 7

g) a =, where a = 0 ad a = a + 0 for every h) b = where b is the -th digit after the decimal poit i the decimal expasio of π i) + ) ), where is a fixed real costat.. Compute the followig series, if they coverge. a) b) c) =0 = e ) = + 6) + 7) d) + + + 9 + 7 6 +... e) f) g) 0 87 + 00 ) 000 = 6 + =0. State whether the followig series coverge or diverge. Carefully justify each aswer. a) l ) 0 =0! b) =0 c) e d) e) f) g) 4 l ) + + + 4) = =

4 h) i) j) )!) )! = 00 + =0 ) = 8 + ) 4. Determie whether the followig series diverge, coverge absolutely, or coverge coditioally. ) a) b) c) d) e) f) g) = = = ) si ) + 0 = = )! ) 4 + 4 4/ + /

Aswer Key Disclaimer: This ey was writte quicly ad may cotai errors or typos! Please let me ow if you have detected oe.. a. b. π c. diverges d. diverges e. 0 f. 0 g. h. diverges sice π is irratioal! i. e. a. e e b. 0 c. 7 d. 6 e. diverges f. 4 g. 6. a. diverges Sugg: Divergece Test) b. coverges Sugg: Ratio Test) c. coverges Sugg: Itegral Test) d. coverges Sugg: Direct Compariso Test) e. coverges Sugg: Root Test) f. diverges Sugg: Limit Compariso Test) g. diverges Sugg: Limit Compariso Test) h. coverges Sugg: Ratio Test) i. coverges Sugg: Root Test) j. coverges Sugg: Root Test). coverges Sugg: Direct Compariso Test) 4. a. coverges coditioally Sugg: AST, ad p-series) b. coverges absolutely p-series) c. coverges absolutely Sugg: DCT) d. diverges Terms do t coverge to 0!) e. coverges absolutely geometric series) f. coverges absolutely Sugg: Ratio Test) g. coverges coditioally Sugg: AST, ad LCT with p-series)