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JH

For what values of $p$ does the series $\sum_{n = 2}^{\infty} 1/(n^P \ln n)$ converge?

converges for $p>1$

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Anna Marie V.

Campbell University

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Samuel H.

University of Nottingham

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Let's find the P values for which this series will converge. So will show that we actually need P to be strictly larger than one. So first, let's consider the case when piece bigger than one. So let's actually show that we have convergence in this case. So the Siri's I'LL go ahead and rewrite this Siri's Let me just pull off this first term by plugging in and equals two one over to the pl in two and then I have three all the way to infinity. Now the reason I'm doing this is because if N is bigger than an equal to three natural law government is bigger than one. So one over np natural log of end is less than one over and to the P. So the right hand side is large because has a smaller denominator. So let's rewrite this. Now we use upper bound. It's one over into the P and we know that this converges Bye, assumption. But so here this is by the pee test and here he is bigger than one. So recall the pee test. This is from a section eleven point three and here this is just a number so adding a number to a conversion Siri's is the same. It's just adding to roll numbers. So this right hand side just isn't real number. So converges. And then we're using the comparison test because we haven't upper bound here. So by the comparison test, the theories that we started with can also converges. So that's comparison test. Now let's show in the other two cases, P equals one and pee less than one that we have diversions. So I'm running out of room here, so I'll go to the next page for this. So hear people's one and we're looking at the Siri's from two to infinity one over end to the one because P is one, and for this, we can use probably should use the integral test. So here this tells us that we want to consider this function of of ex Where s EV en equals one over and Ellen X. So the inner girl test says we just look at the integral from to infinity and let me go back and fix that mistake That should be an integral Tess's. We have this integral one over x l n x, and we'LL show that this integral diverges. So here, let's do a U substitution. Then we have Ellen of two natural log of infinity is still infinity. So what I actually do here before I do the use of Let me go ahead and write. This is a limit, so I'm not doing the use of yet. Then my next step is to go ahead and do the use of. So now I have to switch my limits of integration. And then I have won over you. Do you? So that's just natural log Absolute value. You from Ellen too, To El Ante. So I have natural log. Absolute value. LMT minus natural log Absolute value. Ellen too. Now, as we take the limit, we know that the natural law goes to infinity. So that means this absolute value goes to infinity and therefore this outer wall. Either of them will also go to infinity. So this limit equals infinity. This is diverges. So by the interval test our Siri's and equals two to infinity One over and p national lot of end diverges when P equals one. We have one more case to consider. So going on to the next page, the final case he is less than one. In that case, this will be larger than or equal to the sum of one over and Ellen and so here. All I did was replace P with one here, and P is less than or equal to n if P is less than one. So the right hand side has a larger denominator, and that's why the fraction is a hole that smaller. But we just showed there's some diverges in the previous case when P equals one. Therefore, by the comparison test, this larger Siri's must also diverge. Shin No, therefore, and to summarize this on ly ca merges converges on Lee when P is nature than one, and that's our final answer.

JH

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Anna Marie V.

Campbell University

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Samuel H.

University of Nottingham

Lectures

Join Bootcamp