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

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

10.6 ALTERNATING SERIES

6.3 Testing Series With Positive Terms

INFINITE SEQUENCES AND SERIES

10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

Calculus II - Problem Drill 21: Power Series, Taylor and Maclaurin Polynomial Series

9.3 The INTEGRAL TEST; p-series

JANE PROFESSOR WW Prob Lib1 Summer 2000

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

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)

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

Solutions to Tutorial 5 (Week 6)

Math 113 Exam 3 Practice

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

Quiz 5 Answers MATH 141

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

Testing for Convergence

Ma 530 Introduction to Power Series

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.

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

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

Infinite Sequences and Series

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

Series Solutions (BC only)

7 Sequences of real numbers

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.

Part I: Covers Sequence through Series Comparison Tests

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

The Interval of Convergence for a Power Series Examples

Infinite Sequence and Series

CHAPTER 10 INFINITE SEQUENCES AND SERIES

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

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

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

Series: Infinite Sums

Section 11.8: Power Series

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

Quiz No. 1. ln n n. 1. Define: an infinite sequence A function whose domain is N 2. Define: a convergent sequence A sequence that has a limit

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

MATH 31B: MIDTERM 2 REVIEW

MAT1026 Calculus II Basic Convergence Tests for Series

INFINITE SEQUENCES AND SERIES

Chapter 6: Numerical Series

THE INTEGRAL TEST AND ESTIMATES OF SUMS

Chapter 10: Power Series

Solutions to Practice Midterms. Practice Midterm 1

MA131 - Analysis 1. Workbook 3 Sequences II

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

MA131 - Analysis 1. Workbook 9 Series III

Ma 530 Infinite Series I

Chapter 7: Numerical Series

Series III. Chapter Alternating Series

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

Practice Test Problems for Test IV, with Solutions

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

Math 181, Solutions to Review for Exam #2 Question 1: True/False. Determine whether the following statements about a series are True or False.

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 +

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

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

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.

Solutions to Tutorial 3 (Week 4)

11.6 Absolute Convergence and the Ratio and Root Tests

Math 106 Fall 2014 Exam 3.1 December 10, 2014

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

n n 2 n n + 1 +

Sequences. Notation. Convergence of a Sequence

5.6 Absolute Convergence and The Ratio and Root Tests

9/24/13 Section 8.1: Sequences

5 Sequences and Series

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

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

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

Chapter 6 Infinite Series

Math 106 Fall 2014 Exam 3.2 December 10, 2014

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

M17 MAT25-21 HOMEWORK 5 SOLUTIONS

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.

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

Not for reproduction

Math 210A Homework 1

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

Additional Notes on Power Series

Roberto s Notes on Infinite Series Chapter 1: Sequences and series Section 3. Geometric series

Math 112 Fall 2018 Lab 8

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

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

Section 5.5. Infinite Series: The Ratio Test

1 Introduction to Sequences and Series, Part V

Math 163 REVIEW EXAM 3: SOLUTIONS

Strategy for Testing Series

9.3 Power Series: Taylor & Maclaurin Series

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

Math 113 Exam 4 Practice

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

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

Math 113 Exam 3 Practice

Solutions to Homework 7

Math 116 Second Midterm November 13, 2017

Transcription:

Calculus II - Problem Solvig Drill 8: Sequeces, Series, ad Covergece Questio No. of 0. Fid the first four terms of the sequece whose geeral term is give by a ( ) : Questio #0 (A) (B) (C) (D) (E) 8,,, 4 8,,, 4 8,,, 4 8,,, 4 8 4,,, A. Icorrect! The sigs are the terms are icorrect. B. Correct! This is the correct aswer foud by pluggig,,, & 4 ito the geeral formula. Feedback o Each Aswer C. Icorrect! The sigs are the terms are icorrect. D. Icorrect! The sigs are the terms are icorrect. E. Icorrect! Plug,,, & 4 ito the geeral term formula. If the geeral term is give by a ( ), the the first four terms are foud by 8 pluggig,,, & 4 ito the formula. We have a, a, a, ad 4 4 a. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. of 0. Fid the first four terms of the sequece whose geeral term is give by a ( ) : Questio #0 (A),, 9, 64 (B),, 9, 64 (C),, 9, 64 (D),, 9, 64 (E),, 9, 64 A. Correct This is the correct aswer foud by pluggig,,, & 4 ito the geeral formula. B. Icorrect! The sigs are the terms are icorrect. Feedback o Each Aswer C. Icorrect! The sigs are the terms are icorrect. D. Icorrect! The sigs are the terms are icorrect. E. Icorrect! The sigs are the terms are icorrect. If the geeral term is give by a ( ), the the first four terms are foud by pluggig,,, & 4 ito the formula. We have a, a, a 9, ad 4 64 a. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. of 0. Fid a geeral formula for the sequece 4 8,,,,... 5 Questio #0 (A) (B) (C) (D) a a a + + a (E) Ca t be determied A. Icorrect! Try to recogize a patter. It might help to rewrite / as 6/4. B. Icorrect! Try to recogize a patter. It might help to rewrite / as 6/4. Feedback o Each Aswer C. Correct! This is the correct aswer. It is foud by recogizig the patter. D. Icorrect! Try to recogize a patter. It might help to rewrite / as 6/4. E. Icorrect! Oe of the give aswers is correct. Please try agai. We begi by rewritig the sequece 4 8,,,,... 5 as 4 6 8,,,,... 4 5 We may ow recogize the umerator as ad the deomiator as +. Therefore, the geeral term is a. + Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 4 of 0 4. Cosider the sequece with geeral term sequece coverge to? a +. What value does this Questio #04 (A) + (B) (C) (D) (E) + A. Icorrect! Take the limit as approaches ifiity. B. Icorrect! Take the limit as approaches ifiity. Feedback o Each Aswer C. Correct! This is the correct aswer. It is foud by takig the limit as approaches ifiity. D. Icorrect! Take the limit as approaches ifiity. E. Icorrect! Take the limit as approaches ifiity. a + lim a lim. + We fid the value that We have coverges to by takig the limit as approaches. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 5 of 0 5. The sequece with geeral term a is: Questio #05 (A) Bouded (B) Mootoic (C) Bouded ad mootoic (D) Not mootoic ad ot bouded (E) Noe of these. A. Icorrect! The sequece is ot bouded. B. Correct! This is the correct aswer. The sequece is mootoic, but ot bouded. Feedback o Each Aswer C. Icorrect! The sequece is ot bouded. D. Icorrect! The sequece is mootoic. E. Icorrect! Oe of the give choices is correct. Please try agai. Give a, it is clear that the sequece is ot bouded sice approaches as gets larger ad larger. Sice this is a strictly icreasig sequece, it is mootoic. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 6 of 0 6. a is called Questio #06 (A) A ifiite series (B) A ifiite sequece (C) A geeral term (D) A fiite series (E) A fiite sequece A. Correct! This is the correct aswer. This is a ifiite series. B. Icorrect! A sequece is a list of terms, ot the sum. Feedback o Each Aswer choice C. Icorrect! The geeral term is part of this expressio, but the overall expressio is a ifiite series. D. Icorrect! A fiite series would ot have ifiity as the upper limit. E. Icorrect! A sequece is a list of terms, ot the sum. Expressios i the form a are called ifiite series. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 7 of 0 7. What coclusio ca be made about the covergece or divergece of usig the divergece test? Questio #07 (A) Coverge (B) Both coverge ad diverge (C) Diverge (D) Neither coverge or diverge (E) No coclusio ca be made A. Icorrect! Take the limit of the geeral term a approaches ifiity. B. Icorrect! It is ot possible to both coverge ad diverge at the same time. Feedback o Each Aswer C. Correct! This is the correct aswer. Sice the limit of () as approaches ifiity is ot zero, the series diverges. D. Icorrect! The series must either coverge or diverge. E. Icorrect! Take the limit of the geeral term a approaches ifiity. Give, we fid that lim( ). By the divergece test, the series diverges. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 8 of 0 8. What coclusio ca be made about the covergece or divergece of usig the divergece test? Questio #08 (A) Coverge (B) Both coverge ad diverge (C) Diverge (D) Neither coverge or diverge (E) No coclusio ca be made A. Icorrect! The series does ot coverge. Remember that just because / has a limitig value of 0, this does ot guaratee covergece. B. Icorrect! It is ot possible to both coverge ad diverge at the same time. A Feedback o Each Aswer C. Icorrect! The series actually does diverge. However, this caot be determied based o the divergece test. D. Icorrect! The series must either coverge or diverge. E. Correct! This is the correct aswer. Based o this tutorial, eve though the limitig value of / is 0, o coclusio ca be made about the covergece without further aalysis. Give, we have lim 0. However, this does ot guaratee covergece accordig to the divergece test. Therefore, o coclusio ca be made based o the divergece test. This series is actually a special case of a ifiite series called the Harmoic series ad is kow to diverge. Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 9 of 0 9. Fid the value of (A) i 0 i Questio #09 (B) (C) (D) (E) The series does ot coverge A. Correct! This is the correct aswer. Sice the series is a coverget geometric series, its sum ca be foud easily. B. Icorrect! The series is a covergig geometric series. A Feedback o Each Aswer C. Icorrect! The series is a covergig geometric series. D. Icorrect! The series is a covergig geometric series. E. Icorrect! The series does coverge. It is a geometric series. Note that i 0 i is a geometric series with a ad a r that the series coverges to. r. Sice < r <, we kow Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved

Questio No. 0 of 0 Questio #0 0. Fid the value of (A) 5 (B) 0 (C) i 0 i 5 (D) 5 (E) The series does ot coverge A. Icorrect! The series is a covergig geometric series. Feedback o Each Aswer B. Correct! This is the correct aswer. Sice the series is a coverget geometric series, its sum ca be foud easily. C. Icorrect! The series is a covergig geometric series. D. Icorrect! The series is a covergig geometric series. E. Icorrect! The series does coverge. It is a geometric series. Note that i 0 i 5 is a geometric series with a 5 ad a 5 0 r kow that the series coverges to. r. Sice < r <, we Solutio RapidLearigCeter.com Rapid Learig Ic. All Rights Reserved