anonymous
  • anonymous
I have a series convergence question: Find the values of the parameter p for which the following series converges:
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sum_{k=2}^{\infty} \frac{ 5 }{ (4\ln(\ln k))^{p}) }\]
myininaya
  • myininaya
hmmm...what tests have you studied?
anonymous
  • anonymous
divergence, ratio, root, limit comparison...

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myininaya
  • myininaya
for some reason I'm thinking we should try a comparison test
anonymous
  • anonymous
hmm...i'll try that
myininaya
  • myininaya
\[\log(\log(k))1 \]
jtvatsim
  • jtvatsim
I have a hunch that the ratio test might be helpful... not sure yet, working on the result.
jtvatsim
  • jtvatsim
just to clarify the entire bottom is being raised to the pth power correct?
anonymous
  • anonymous
yes
jtvatsim
  • jtvatsim
k
anonymous
  • anonymous
im not sure...my results for comparison test says the series diverges...=/ stuck
myininaya
  • myininaya
for which p?
myininaya
  • myininaya
I don't think it should diverge for all p
myininaya
  • myininaya
could be wrong though
myininaya
  • myininaya
\[\frac{1}{k}< \frac{1}{\ln(\ln(k))} \\ \text{ if } p>0 \text{ then we have } \\ \frac{1}{k^p}< \frac{1}{(\ln(\ln(k)))^p} \\ \] so in this comparison the only thing we can see if for p>0 we have divergence
myininaya
  • myininaya
\[\int\limits_{2}^{\infty} \frac{1}{k^p} dk =\int\limits_2^\infty k^{-p} dk \\ \text{ assume } p \neq 1 \text{ then we have } \frac{k^{-p+1}}{-p+1}|_2^{\infty} \\ =\lim_{z \rightarrow \infty}(\frac{z^{1-p}}{1-p})-\frac{2^{1-p}}{1-p} \\ \\ \text{ hmm if } 1-p>0 \text{ then the \integral diverges } \\ \text{ and if } 1-p<0 \text{ then \integral converges } \] 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 1-p>0 we have the series diverges
myininaya
  • myininaya
oops and we had p>0 above so actually p is between 0 and 1 there
myininaya
  • myininaya
but still need to look at other p cases
myininaya
  • myininaya
for example what happens at p=0 and p=1 and everything outside the (0,1) interval
myininaya
  • myininaya
so the only thing I have shown is this: \[\text{ Since where } 0 \frac{1}{k^p} \text{ then } \sum_{n=2}^{\infty} \frac{1}{(\ln(\ln(k))^p} \\ \text{ diverges again for } 0
myininaya
  • myininaya
won't (not want)
myininaya
  • myininaya
if p=1 above.. \[\int\limits_{2}^{\infty}\frac{1}{k} dk=\ln|k|_2^\infty \] this still diverges
myininaya
  • myininaya
so you could say 0
myininaya
  • myininaya
so I think we need another sequence for comparision for p>1 we also need to consider p=0 and p<0 brb...
jtvatsim
  • jtvatsim
Clearly, p < 0 diverges as this will bring the denominator on top... and p = 0 gives an infinite series of 5's also divergent.
myininaya
  • myininaya
true above the p=0 thing ... and yes for p<0 we should have divergence
myininaya
  • myininaya
so @cstarxq looks like you were correct for all p this thing diverges
myininaya
  • myininaya
wait we didn't consider p>1 yet
anonymous
  • anonymous
thanks, i just finished this hw question online, the series diverges for all values of p
myininaya
  • myininaya
oh ok cool
myininaya
  • myininaya
was it multiple choice?
anonymous
  • anonymous
with fill-ins
myininaya
  • myininaya
oh ok cool stuff
myininaya
  • myininaya
still trying to think what we can do for p>1
myininaya
  • myininaya
I know you already done it but I might play with a bit more
anonymous
  • anonymous
:) that's cool
myininaya
  • myininaya
@jtvatsim do you have anything for p>1 ?
jtvatsim
  • jtvatsim
Still working on it... Mathematica says to use a comparison test, but doesn't specify what to compare it to.
jtvatsim
  • jtvatsim
I 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
  • anonymous
i actually tried that, getting rid of the logs...but got stuck on that
anonymous
  • anonymous
for limit comparison test, compare it to 1/k?
jtvatsim
  • jtvatsim
tried that... lim comp fails... goes to infinity.
jtvatsim
  • jtvatsim
Ah... might have something.
jtvatsim
  • jtvatsim
This 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
  • myininaya
are you saying: \[\frac{1}{(\ln(k))^e}=\frac{1}{k}?\]
jtvatsim
  • jtvatsim
dang it... lol
jtvatsim
  • jtvatsim
Assume (ln k)^e = 1. Then the result follows. LOL. That's called seeing what you want to. :)
myininaya
  • myininaya
I totally get that. I do that sometimes.
myininaya
  • myininaya
Then everyone else is like math is really magic.
jtvatsim
  • jtvatsim
Yep, 0 = 1 implies anything.
myininaya
  • myininaya
I'm kind of stumped
jtvatsim
  • jtvatsim
Yeah, this is a tough question. There must be some trick involved here.
anonymous
  • anonymous
It 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
  • jtvatsim
Really? Cuz, that didn't work for me. The way my math is going right now though, I guess that may be true...
jtvatsim
  • jtvatsim
Well, I am done for tonight. Have a good one everyone! :)
anonymous
  • anonymous
goodnight, thanks for helping!~
myininaya
  • myininaya
goodnight @jtvatsim
myininaya
  • myininaya
@cstarxq sadly I guess I'm not picking the right series for comparison I give up for tonight :(
myininaya
  • myininaya
oh i think i see I guess i forgot all the parts of the comparison limit test
myininaya
  • myininaya
https://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
  • myininaya
http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx I was using pauls notes and did not see this for the limit comparison test.
myininaya
  • myininaya
That was my guide until I did some more searching.
myininaya
  • myininaya
and found that pdf
anonymous
  • anonymous
@myininaya oh wow, that pdf will be really helpful for studying for my upcoming exam. Thanks for sharing!

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