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

Similar documents
Infinite Sequence and Series

Section 11.8: Power Series

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

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

Section 5.5. Infinite Series: The Ratio Test

Chapter 6 Infinite Series

Ma 530 Introduction to Power Series

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

INFINITE SEQUENCES AND SERIES

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

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

Ma 530 Infinite Series I

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

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

Infinite Sequences and 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 Introduction to Sequences and Series, Part V

SCORE. Exam 2. MA 114 Exam 2 Fall 2016


Section 7 Fundamentals of Sequences and Series

Sequences, Series, and All That

6.3 Testing Series With Positive Terms

Math 132, Fall 2009 Exam 2: Solutions

n n 2 n n + 1 +

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

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

CHAPTER 1 SEQUENCES AND INFINITE SERIES

2.4.2 A Theorem About Absolutely Convergent Series

The Interval of Convergence for a Power Series Examples

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

MATH 2300 review problems for Exam 2

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.

Section 1.4. Power Series

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 do NOT write in this box. Multiple Choice. Total

Mathematics 116 HWK 21 Solutions 8.2 p580

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

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

Math 25 Solutions to practice problems

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

Chapter 7: Numerical Series

SUMMARY OF SEQUENCES AND SERIES

JANE PROFESSOR WW Prob Lib1 Summer 2000

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

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

Created by T. Madas SERIES. Created by T. Madas

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)

(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

10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

INFINITE SEQUENCES AND SERIES

11.6 Absolute Convergence and the Ratio and Root Tests

MTH 246 TEST 3 April 4, 2014

CHAPTER 10 INFINITE SEQUENCES AND SERIES

9.3 The INTEGRAL TEST; p-series

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

Unit 6: Sequences and Series

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

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

5.6 Absolute Convergence and The Ratio and Root Tests

Chapter 7 Infinite Series

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

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

7 Sequences of real numbers

Chapter 6: Numerical Series

MATH 312 Midterm I(Spring 2015)

10.6 ALTERNATING SERIES

MAT1026 Calculus II Basic Convergence Tests for Series

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

Math 113 Exam 3 Practice

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.

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

is also known as the general term of the sequence

Proposition 2.1. There are an infinite number of primes of the form p = 4n 1. Proof. Suppose there are only a finite number of such primes, say

AP Calculus Chapter 9: Infinite Series

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

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

Math 106 Fall 2014 Exam 3.2 December 10, 2014

MA131 - Analysis 1. Workbook 7 Series I

Math 113 Exam 4 Practice

M17 MAT25-21 HOMEWORK 5 SOLUTIONS

Testing for Convergence

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

THE INTEGRAL TEST AND ESTIMATES OF SUMS

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

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

5 Sequences and Series

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

10.2 Infinite Series Contemporary Calculus 1

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

Series: Infinite Sums

Solutions to Math 347 Practice Problems for the final

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

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

Chapter 10: Power Series

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

MATH 2300 review problems for Exam 2

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

Review of Sections

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

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

Transcription:

Quiz No.. Defie: a ifiite sequece A fuctio whose domai is N 2. Defie: a coverget sequece A sequece that has a limit 3. Is this sequece coverget? Why or why ot? l Yes, it is coverget sice L=0 by LHR.

INFINITE SERIES

Questio. Ca Joh ad Mark cross the ever-edig bridge?

Questio. Let x be the legth of the bridge. x x x x x 2 4 8 6 32... Note that this expressio ca also be writte as x 2

Questio 2. Ca we color the box completely?

Questio 2. 2 4 8 6 32... Note that this expressio ca also be writte as 2

Questio 3. Do you kow the sum? 2 4 8 6 32... Aswer:

Questio 4. Recall the sequece defied by. g 2 3 4 5 6 7 Now, what ca you say about...

Questio 4. WALANG SUM? MAY SUM -? 0? /2?? depede sa last term?

Questio 4.... NO, it does ot have a sum. Note that this expressio ca also be writte as

.3 INFINITE SERIES of CONSTANT TERMS

Defiitio. Let u be a sequece of real umbers ad s u u u u 2 3... The the sequece series. NOTATION: s s, is called a ifiite u

Defiitio. I a ifiite series, u u, u 2,..., u,... s, s 2,..., s,... s are called the terms of the ifiite series are called the partial sums of the ifiite series is the sequece of partial sums defiig the ifiite series

Example. Cosider 2 The first four terms of the series are u 2 u 2 3 4 u u 4 8 6 The first four partial sums of the series are s 7 s 3 2 2 4 8 8 3 s 2 2 4 4 5 s 4 2 4 8 6 6

Remarks: If s is the sequece of partial sums defiig the ifiite series The for 2, s s u. Our mai cocer o ifiite series is to determie whether the series coverges or ot. u

Defiitios. Cosider a ifiite series ad be the sequece of partial sums defiig the ifiite series. lim If exists ad is equal to, The: s u is coverget u S s S is the sum of the ifiite series

Defiitios. If lim s does ot exist, The: u is diverget ad it does ot have a sum.

Ca you add a ifiite umber of terms ad have a fiite sum?...... 2 2 4 8 2

Prove....... 2 2 4 8 2 PROOF. s... 2 4 8 2 2 s... 2 4 8 2 2 2 s s 2 2 2 s 2 2 2

PROOF. (cot.) Now, s 2 2 2 s 2 0 lim s lim 2 Thus, 2 is coverget ad its sum is.

Example 2. Cosider 3 32 Let u 332 Recall that 3 3 2 33 33 2

Example 2. (cot.) Now, 3 32 s u u u... u u 2 3 s 3 2 3 5 3 5 38 3 8 3... 3 3 4 3 3 3 3 3 3 2 A collapsig/telescopig series

Example 2. (cot.) So, s 3 2 3 3 2 0 lim s lim 6 3 3 2 6 Thus, 3 32 is coverget ad its sum is. 6

Theorem. (th term test) lim u 0 If u is coverget, the. PROOF. Suppose lim u is coverget with sum S. The. s S s s u Recall: Thus, lim u lim s s S S 0

Remark: (th term test for divergece) lim u 0 If, the u is diverget. BUT lim u 0 If, the u is either diverget or coverget.

Examples. Is the series diverget? 3 2 2 2 2 5 2 3 l 5 e si 2

Theorem. s If u is coverget ad is the sequece of partial sums defiig the series, the 0 for each, there exists a umber such R T that if ad are atural umbers such that N R N ad T N, the s s. R T

Theorem. The harmoic series is diverget....... 2 3 PROOF. Here we use the previous theorem with s s 2 R 2, T... 2 3...... 2 3 2

PROOF. (cot.) s 2 s... 2... 2... 2 2 2 2 That is, o umber s s 2 Thus, N exists such that whe. 2 is diverget.

Alterative Proof 2 2 2...... 2 3 2 3 4 4 2 4 5 6 7 8 8 2 8... 9 0 5 6 6 2 etc. Thus, k... 2 3 k 2 2 k lim diverges. k 2 is diverget.

Theorem. The geometric series ar a r a 0 a r r where ad are costats ad.. coverges to if ; 2. diverges if r. PROOF. 2 s a ar ar... ar 2 rs ar ar... ar ar s rs a ar

PROOF. (cot.) s r a r s rs a ar s If a r r a r lim s lim r r, the lim s. a r Now, If r, the lim u lim ar 0. Thus, theorem holds.

Examples. Determie if the geometric series is coverget. If it does, fid its sum. 4 5 2 3 4 3 3 2 7 2 5 2 e

Coverget Diverget u where lim s exists u where lim u 0 f f k k ar, r ar, r

Quiz No. 2 coverget or diverget? Briefly explai why.. 2. 3 5 3. 2 sec

4 Theorems about Ifiite Series. Cosider two ifiite series a ad b.. If they differ oly i a fiite umber of terms, the either both series coverge or both diverge.

4 Theorems about Ifiite Series. 2a. If the series a is coverget ad its sum S is, the the series ca is also coverget ad its sum is costat. c cs for each

4 Theorems about Ifiite Series. 2b. If the series a is diverget, the the series da ozero costat. is also diverget for each d

4 Theorems about Ifiite Series. 3. The sum or differece of two coverget series is also coverget. 4. The sum or differece of a coverget series ad a diverget series is diverget.

Examples. Determie whether the series is coverget or diverget. Explai why. 3 2 2 2 3 2 7 5 3 3 e

WARNING!!! The terms of a coverget series ca be grouped i ay way (provided that the order of the terms is maitaied), ad the ew series will coverge with the same sum as the origial series.

END