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# For what value of $p$ is each series convergent?$\displaystyle \sum_{n = 2}^{\infty} ( - 1)^{n-1} \frac {(\ln n)^P}{n}$

## $p$ can be any real number

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

Let's find the values of P for which the syriza's conversion. So here we see that this Siri's is alternating. So let's just go ahead and not make any assumptions yet about Pete. But it could be any real number. And let's see how far we can go with the alternating series test. Perhaps this will converse for any number of pee. So here we defined the end to just be the positive part of this and term here. So issues natural log of end to the P over. And so here. This is bigger than early or actually here. This will actually just being bigger than zero if and is bigger than or equal to two. That's the assumption over here. So actually, I take that back. So here we want and to be bigger than three. All that matters is that it's eventually true. Doesn't have to be true for every end. But there has to be a point in which we hit a number in this case three and every number. After that, it's positive. So then that's our BM. That's the first condition is that it has to be positive. The second condition is that we need the limit as n goes to infinity of being to be zero. So that's natural log of end to the P over end and here what I would do is if I want to use Lopez's house rule. I'Ll just use x So here by low Patel this limit So here really depends on peace off p zero. So first, if if peace equals is you know or even less than zero, then the limit well equal zero. Otherwise it peace positive then it's not so obvious. We have infinity over infinity Then the limit is of the form infinity over infinity. And this is where we use Lopez house rule So we would take the derivative of the numerator And then we get Exxon the denominator from the General And then we take the derivative of the denominator ex specious one. And if we could, then we could look at T minus one. If p minus one is less than or equal to zero, the limit is is equal to zero. Otherwise it appears bigger. P minus one is bigger than zero. I want to use Lopez house rule again and we'Ll continue to keep subtracting one from the P Eventually this will be zero or negative and then we'Ll still have an excellent bottom from the chain rule and this limit will end up being zero. So here I'll summarize that limit of being equal zero and the last condition to check is that BN is eventually decreasing. So let's go to the next page. This was our b bien. So the third condition in the final one for altering Siri's is that the sequence is decreasing eventually, Not not necessarily for all in so there. To show this, we could define this function f and based on the end. And then I know that if that prime is negative, that's equivalent Teo being decreasing. So this is what we're gonna put a question mark here because that's what we want to check. F is decreasing That's equivalent to the sequence being decreasing so we can use this method a line in the book. Teo, check whether beings decreasing witches have to take the derivative of this eh function. So I used the question rule. So here you'll get P l n x if he minus one over x ten times the denominator and then minus l a next to the P times one over denominators, Claire. So this Khun simplify you pull out a Ellen to the P minus one power. So here, if X is bigger than her equal to three, then Ellen of X is positive. I know the denominators positive. The only question is is whether P minus Elena X is positive. If this thing is negative, this means that the entire fraction is negative. And this will be true when piers less than Ellen X, which is equivalent to even the pee less than X. So this is eventually true when X increases so if and is bigger than me to the P, then be n plus one is less than or equal to be in. So this bee in sequence is eventually decreasing. And that's what we wanted increasing. And let me summarize on the last page here by the alternating Siri's test, the Siri's convergence. And now we tio answer the original question. Since we made no assumptions about P, we've shown that the Siri's converges for Alti converges for any really number team, and that's our final answer

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp