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The terms of a series are defined recursively by …

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Problem 38 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}{n \ln n} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 6

Absolute Convergence and the Ratio and Root Tests

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Video Transcript

Let's first check whether this series is absolutely convergent. So to do that instead of looking at these terms here, we replaced them with absolute value. So absolute value of negative onto the end is just one and everything else stays the same because then in natural log of end or both, positive effect is bigger than or equal to two. And excuse me here, This should not be a two. There should be an end. Sorry about that. So now for this, this series here, I would use the integral test so we would look at the integral to to infinity one over X natural log of X. The reason I would do an overall test here is because if you replace and with exes, this could be integrated using a U substitution. So before that, let's bring this over here. Let me rewrite this as an improper, integral and then I would go ahead and do a use up here. Let's try. U equals natural log of X de was one over X DX, so we'll keep that limit. The limit will stay until after we evaluate the integral. So now, using our use of you could take these X values over here and plug them into this formula to find your you bounds of integration. And then the integral becomes one over you. Do you natural log Absolute value of you. U goes from l N two to lnk and as we take the limit, natural log of K goes to infinity And so therefore the natural log of this will also go to infinity whereas we'll have over here natural log of natural log of two which is not infinite. It's just a number. So when we do infinity minus this number, we still get infinity so the integral does not converge. This means that this series over here not the original but the modified one This one diverges So we can say that our original series is not absolutely convergent But then it's possible that it's still conditionally convergent. So let's go ahead and check this. Let me go on to the next page. Yeah, yeah, So the question now is whether it's conditionally convergent. So here, let's try. We see that this is an alternating series so let's try the alternating series test. So here we can define bn to just be won over end natural out of end. So we're again. We're taking the absolute value here and then notice that this is positive Sense and a natural log event are both positive if n is bigger than or equal to two, which is the case in our problem. So that's one condition of the alternating series test. The next one, you need the limit of the BN to be zero. In our case, both of the factors in the denominator go to infinity Absolutely. As we take the limit. So the fraction becomes one over infinity, which is zero. And finally we need to check that the BNS are decreasing. So how are we going to verify that? So one way is to just convince yourself that this equation is true in our case and this is equivalent to and natural log of end, less than or equal to n plus one natural log of n plus one. And the latest inequality is true because N is less than n plus one. A natural log event is less than natural log event plus one. Now, since these are all equivalent, that means that this is true. Which means that the B m is a decreasing sequence. So we conclude that the series converges by a S t. And since we already showed it was an absolutely conversion, the answer is that it's conditionally convergent good.

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Problem

The terms of a series are defined recursively by …

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 11

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

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

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals