Problem

Use any test to determine whether the series is a…

02:42

Question

Answered step-by-step

Problem 31 Hard Difficulty

Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$ \displaystyle \sum_{n = 2}^{\infty} \frac {( - 1)^n}{\ln n} $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by J Hardin

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 6

Absolute Convergence and the Ratio and Root Tests

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 11

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Video Transcript

let's first determine whether the series is absolutely conversion. So we'LL take the absolute value of this term here, and that leaves us with just one over natural log of end. So the question now is whether this new Siri's converges. And so here I would use the comparison test we have won over Natural log of Team. This is excuse me. Natural albumen is bigger than or equal to one over end. And the reason for this is that N is bigger than are equal to, but really bigger than natural log of end and in our case and his two or larger. So this justifies this over here, and we know that the Siri's one over and diversions this's the harmonic series or ego you could also use. The Pee test with P equals one. So since our term up here was bigger than or equal to end, then by the comparison test our series starting at two won over national log of n diverges. So that's using comparison. So here we conclude that it's not absolutely commercial now will determine whether it's conditionally conversion. So this is asking whether or not the original Siri's over here, convergence So you see that this is alternating Siri's. So that suggests we should go ahead and use the alternating Siri's test. Let me go to the next page. So am I. Some here from to the infinity. Negative one and natural law again. Slow fuse alternating Siri's test s t. So here we defined B end to just be the absolute value of this term over here. So that's one over natural log of end and notice that this is bigger than or equal to zero. If Anne is bigger than or equal to two coming from over here the lower bound of the sun. So we have to check a few conditions here. We need that the beings are decreasing. We'LL check that later. Let's first just check that the limit is zero and the denominator goes to infinity. So that limit is zero. We need that to be answered, decreasing. So we would want to show that this is less than or equal to this and this is equivalent to one over natural log and plus one and then we have natural log in. And this is a true statement since we know that natural log of end is less than natural log and plus one. And if you're not convinced why this is true, you could exponentially in each side and you'Ll end up with this. And we definitely know this is true. So we have that the beings are positive that they're limited zero and that the sequence is decreasing. So we we conclude that the Siri's is conditionally conversion.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem