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Find the values of $ p $ for which the series is convergent.

$ \displaystyle \sum_{n = 3}^{\infty} \frac {{1}}{{ n \ln n [\ln (\ln n)]^P}} $

$\mathrm{p} > 1$

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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Boston College

here. We'd like to know which P values make this Siri's conversion. So this is a rough looking some here, so but try to use the integral test. So here I would look at the function F banks, sir. Just replace all the ends with exes. So here, if you take any ex larger than or equal to three, then this will imply that X Ellen eggs and then Ellen Ellen eggs. Also, we're all positive, and you can check Gorgo the calculator and see, that's the smallest value of Ellen Ellen. Because once X increases, this also increases. So this double log also increases a sex increases. Okay, so that guarantees that FX is not negative part of the integral, tense requirement. Also, we need that f of X is decreasing Now. The easiest way to see this is that as Ex gets larger, all these terms in the denominator also get larger. Another way to see this with more difficult. But you could compute the derivative and then show it's negative if X is bigger than or equal to three. And that's another way now we've got. Now we're in a condition. Since these two are satisfied these are the hypotheses for the animal tests. So those were satisfied. So that means instead of looking at the song, we can look at the following natural log of natural log of three to infinity. And then now we have a fix. And what's the best way to compute this? There's probably several ways, but perhaps the easiest substitution here that one could make, which is beat to take you to be the double log. Because if we do this, this term here in the bottom just becomes you to the P. And also, if you take the derivative differential here and use the chain rule, then you get the remaining part of the immigrant. So there was a little typo appear when you go back and change this. I did not do the substitution at this point so that Lower Bound should have just been a three. Then after the use of here, now we have because right here that's the X value. But now we have to use the corresponding you value. And the way to do that is you take the Explorer bound, plug it into the U sub, and then that new value becomes your new lower bound and the same could be said for the upper bound infinity because if you take the limit here is exposed to infinity. That also goes for infinity. So now we're just left over with one over U P and by the pee test. This converges if, and only if he's bigger than one. So therefore, by integral test, we conclude that our Siri's converges if and only if peace larger than one.