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anonymous
 one year ago
I have a series convergence question: Find the values of the parameter p for which the following series converges:
anonymous
 one year ago
I have a series convergence question: Find the values of the parameter p for which the following series converges:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=2}^{\infty} \frac{ 5 }{ (4\ln(\ln k))^{p}) }\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2hmmm...what tests have you studied?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0divergence, ratio, root, limit comparison...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2for some reason I'm thinking we should try a comparison test

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\log(\log(k))<k \text{ for } k>1 \]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I have a hunch that the ratio test might be helpful... not sure yet, working on the result.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1just to clarify the entire bottom is being raised to the pth power correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure...my results for comparison test says the series diverges...=/ stuck

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I don't think it should diverge for all p

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2could be wrong though

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{k}< \frac{1}{\ln(\ln(k))} \\ \text{ if } p>0 \text{ then we have } \\ \frac{1}{k^p}< \frac{1}{(\ln(\ln(k)))^p} \\ \] so in this comparison the only thing we can see if for p>0 we have divergence

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{2}^{\infty} \frac{1}{k^p} dk =\int\limits_2^\infty k^{p} dk \\ \text{ assume } p \neq 1 \text{ then we have } \frac{k^{p+1}}{p+1}_2^{\infty} \\ =\lim_{z \rightarrow \infty}(\frac{z^{1p}}{1p})\frac{2^{1p}}{1p} \\ \\ \text{ hmm if } 1p>0 \text{ then the \integral diverges } \\ \text{ and if } 1p<0 \text{ then \integral converges } \] but we can only use the diverging thing since we chose a sequence less than the given sequence for comparison so the only thing we have shown is that for 1p>0 we have the series diverges

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2oops and we had p>0 above so actually p is between 0 and 1 there

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2but still need to look at other p cases

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2for example what happens at p=0 and p=1 and everything outside the (0,1) interval

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so the only thing I have shown is this: \[\text{ Since where } 0<p<1 \text{ the } \int\limits _2^\infty \frac{1}{k^p} dk \text{ diverges then } \\ \sum_{n=2}^{\infty} \frac{1}{k^p} \text{ diverges and since } \\ \text{ for } 0<p<1 \text{ we have } \frac{1}{(\ln(\ln(k))^p}> \frac{1}{k^p} \text{ then } \sum_{n=2}^{\infty} \frac{1}{(\ln(\ln(k))^p} \\ \text{ diverges again for } 0<p<1 \\ \text{ we still need to look at other } p \\ \text{ also any constant value times that want change the divergence except maybe 0 :p}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2if p=1 above.. \[\int\limits_{2}^{\infty}\frac{1}{k} dk=\lnk_2^\infty \] this still diverges

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so you could say 0<p<=1 we have divergence

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so I think we need another sequence for comparision for p>1 we also need to consider p=0 and p<0 brb...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Clearly, p < 0 diverges as this will bring the denominator on top... and p = 0 gives an infinite series of 5's also divergent.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2true above the p=0 thing ... and yes for p<0 we should have divergence

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so @cstarxq looks like you were correct for all p this thing diverges

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2wait we didn't consider p>1 yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks, i just finished this hw question online, the series diverges for all values of p

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2was it multiple choice?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2still trying to think what we can do for p>1

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I know you already done it but I might play with a bit more

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2@jtvatsim do you have anything for p>1 ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Still working on it... Mathematica says to use a comparison test, but doesn't specify what to compare it to.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I was wondering if maybe something like 1/ln(ln k)^p > 1/ln(k)^p might be helpful if we somehow express p in terms of an "e" to get rid of the logs...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i actually tried that, getting rid of the logs...but got stuck on that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for limit comparison test, compare it to 1/k?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1tried that... lim comp fails... goes to infinity.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Ah... might have something.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1This should show that the series diverges for p >= e. Given that ln(ln k) < ln k < k for all k > 1. It follows that 1/ln(ln k) > 1/ln k > 1/k for all k > 1. Furthermore, for p > 0 we have, (1/ln(ln k))^p > (1/ln k)^p > (1/k)^p. Focus on the first inequality, namely, (1/ln(ln k))^p > (1/ln k)^p and suppose that p >= e. Then, p = e + r where r >= 0. Then, (1/ln(ln k))^p > (1/ln k)^(e + r) = 1/[(ln k)^e (ln k)^r] = 1/(ln k)^r * 1/k. But this is just a positive constant times a divergent series. Hence, by comparison we have shown that the series diverges for all p >= e. A similar argument shows that the series diverges for 0 < p <= e as well.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2are you saying: \[\frac{1}{(\ln(k))^e}=\frac{1}{k}?\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Assume (ln k)^e = 1. Then the result follows. LOL. That's called seeing what you want to. :)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I totally get that. I do that sometimes.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Then everyone else is like math is really magic.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yep, 0 = 1 implies anything.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, this is a tough question. There must be some trick involved here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It said that the limit comparison for harmonic series is unbounded for all values of p, which means that the series diverges for all values of p... ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Really? Cuz, that didn't work for me. The way my math is going right now though, I guess that may be true...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Well, I am done for tonight. Have a good one everyone! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0goodnight, thanks for helping!~

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2@cstarxq sadly I guess I'm not picking the right series for comparison I give up for tonight :(

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2oh i think i see I guess i forgot all the parts of the comparison limit test

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=0CDIQFjAD&url=http%3A%2F%2Ffaculty.bucks.edu%2Ferickson%2Fmath141%2F141chap9.pdf&ei=AXGTVZqQGsiCsAWszoDgCw&usg=AFQjCNH8TAWi7iA3eNJKKQAQN1hHtM_F7A&sig2=ZbCUCOFfBhLahryJWPpNVA page 19 ... so I was looking at: \[\lim_{k \rightarrow \infty} \frac{1}{\ln(\ln(k))} \div \frac{1}{k} =\infty\] which we can raise both sides by p and we know as you said the harmonic series diverges for all p so the other series also diverges for all p according to page 19 thereom 9.36 part 3

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx I was using pauls notes and did not see this for the limit comparison test.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2That was my guide until I did some more searching.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@myininaya oh wow, that pdf will be really helpful for studying for my upcoming exam. Thanks for sharing!
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