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The Riemann zeta-function $ \zeta $ is defined by…

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Problem 32 Hard Difficulty

Find the values of $ p $ for which the series is convergent.
$ \displaystyle \sum_{n = 1}^{\infty} \frac {\ln n}{n^P} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 3

The Integral Test and Estimates of Sums

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Video Transcript

so here we'd liketo find what P values make this Siri's conversion. So let's apply the integral test so we can define after Becks as Ellen X overexert the P. And here you see Ann is at least one. So we'LL take Ex to be at least one now. To apply the integral tests, we have to make sure that the hypotheses air satisfied. So the first one ethics is positive if X is bigger than one, and we can see this because natural log in exit appealable positive. And if you divide two positives, it's still positive. Further, we need that affects is monitoring decreasing? And perhaps the easiest way to show this message is compute derivative of F and show that it's negative. So here you would just use the potion rule and then show that it's negative. If X is larger than one. And if you have to, you apply abound on P. That's fine, too, but you shouldn't have to hear. Okay, now, using the integral test, we look at the integral and here, let's go ahead into a use of we'LL have to be a little bit careful here because right now and also this one becomes hopes this one over here becomes natural Aga One which is zero and then we have If I write you do you That gives me ln X over x t x But I need extra the p So we have X So we need another excellent p minus one factor here. And here's how we'Ll obtain that so exponentially a both sides of this and then raise both sides of the P minus one And there we go. So we'LL put that instead of writing X to the P minus one in the denominator we'Ll write it in this form over here Now we've ensured that we haven't changed the integral and let me just go ahead and write Those says just put a negative on that p minus one. Now this is the integral that will deal with so maybe write that Okay, so this is an improper integral so we should probably rewrite this is a limit before we integrate. Okay, so well, the older women at the Barre and now we have to integrate this on the inside and here argues integration by parts so using parts here will give us the following and then we have our limits. Zero kay. So then the next step is just the garden plug in those lower and upper limits and zero in K and for you. So that's when you plug in K. Then go ahead and plug in zero the first term. Just become zero so clearly we can see that we should not be taking so far. We are considering Pena equals one. Otherwise thes denominators wouldn't make any sense, so P equals One will deal with that in a moment. But let's not worry about this case right now. So right now we're assuming we're in this case so that all these fractions are defined. Now let's consider two cases, so P is not equal to one. So we have either ppe us than one or bigger than one. So let's assume, for the time being that it's less than one and see if the inner world emerges. So P is less than one. That implies one minus p is positive. And then so now when you take the Limited's cake goes to infinity. This first term right here we'LL go to Well, Kay's going to infinity and then you have a positive number times infinity. So that's also eat to the infinity. And that was just infinity. So that'LL be diversion on the other hand, a piece larger than one, then one minus piece Negative. And this is a good thing because now this becomes infinity. But now we have e to the minus infinity because of the negative, the one minus t being a negative And this limit if you rewrite this So now right is a p minus one p minus one is positive, just won't supply this by a negative and use low Patel's rule Lopez house rule here because it's infinity over affinity and you will get zero here and similarly, this is either the minus infinity. So that will go to zero. And we have a bunch of real numbers that we add. And this will converge no infinities at the very end, so it will converge. So we're mostly done here Now, in the next page will just have to deal with the case where P equals one. So he is one and this case just go back to Original Siri's. Now, this no is bigger than this some here. So perhaps I could be more formal here, So let me go back. So ended The peaches become Zen now was plugging the first few terms Ellen of one zero and so on. And then let me rewrite this as and you'LL see why I'm point pulling out the three in the second. It's because of the following facts. If N is bigger than or equal to three, then natural log of end will be bigger than equals one. So the Siri's is larger than the harmonic series. So here you this diverges, you could either say harmonic. Siri's orjust used to pee Test with P equals one and therefore, by comparison, test and P diversions if P equals one. So, therefore about Wilby showed on the previous page. The original Siri's Comm urges if and only if he's bigger than one.

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Problem

The Riemann zeta-function $ \zeta $ is defined by…

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Watch More Solved Questions in Chapter 11

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Michael Jacobsen

Joseph Lentino

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

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Kayleah Tsai

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals