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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:

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  1. anonymous
    • one year ago
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    \[\sum_{k=2}^{\infty} \frac{ 5 }{ (4\ln(\ln k))^{p}) }\]

  2. myininaya
    • one year ago
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    hmmm...what tests have you studied?

  3. anonymous
    • one year ago
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    divergence, ratio, root, limit comparison...

  4. myininaya
    • one year ago
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    for some reason I'm thinking we should try a comparison test

  5. anonymous
    • one year ago
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    hmm...i'll try that

  6. myininaya
    • one year ago
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    \[\log(\log(k))<k \text{ for } k>1 \]

  7. jtvatsim
    • one year ago
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    I have a hunch that the ratio test might be helpful... not sure yet, working on the result.

  8. jtvatsim
    • one year ago
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    just to clarify the entire bottom is being raised to the pth power correct?

  9. anonymous
    • one year ago
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    yes

  10. jtvatsim
    • one year ago
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    k

  11. anonymous
    • one year ago
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    im not sure...my results for comparison test says the series diverges...=/ stuck

  12. myininaya
    • one year ago
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    for which p?

  13. myininaya
    • one year ago
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    I don't think it should diverge for all p

  14. myininaya
    • one year ago
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    could be wrong though

  15. myininaya
    • one year ago
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    \[\frac{1}{k}< \frac{1}{\ln(\ln(k))} \\ \text{ if } p>0 \text{ then we have } \\ \frac{1}{k^p}< \frac{1}{(\ln(\ln(k)))^p} \\ \] so in this comparison the only thing we can see if for p>0 we have divergence

  16. myininaya
    • one year ago
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    \[\int\limits_{2}^{\infty} \frac{1}{k^p} dk =\int\limits_2^\infty k^{-p} dk \\ \text{ assume } p \neq 1 \text{ then we have } \frac{k^{-p+1}}{-p+1}|_2^{\infty} \\ =\lim_{z \rightarrow \infty}(\frac{z^{1-p}}{1-p})-\frac{2^{1-p}}{1-p} \\ \\ \text{ hmm if } 1-p>0 \text{ then the \integral diverges } \\ \text{ and if } 1-p<0 \text{ then \integral converges } \] 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

  17. myininaya
    • one year ago
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    oops and we had p>0 above so actually p is between 0 and 1 there

  18. myininaya
    • one year ago
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    but still need to look at other p cases

  19. myininaya
    • one year ago
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    for example what happens at p=0 and p=1 and everything outside the (0,1) interval

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

  21. myininaya
    • one year ago
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    won't (not want)

  22. myininaya
    • one year ago
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    if p=1 above.. \[\int\limits_{2}^{\infty}\frac{1}{k} dk=\ln|k|_2^\infty \] this still diverges

  23. myininaya
    • one year ago
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    so you could say 0<p<=1 we have divergence

  24. myininaya
    • one year ago
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    so I think we need another sequence for comparision for p>1 we also need to consider p=0 and p<0 brb...

  25. jtvatsim
    • one year ago
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    Clearly, p < 0 diverges as this will bring the denominator on top... and p = 0 gives an infinite series of 5's also divergent.

  26. myininaya
    • one year ago
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    true above the p=0 thing ... and yes for p<0 we should have divergence

  27. myininaya
    • one year ago
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    so @cstarxq looks like you were correct for all p this thing diverges

  28. myininaya
    • one year ago
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    wait we didn't consider p>1 yet

  29. anonymous
    • one year ago
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    thanks, i just finished this hw question online, the series diverges for all values of p

  30. myininaya
    • one year ago
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    oh ok cool

  31. myininaya
    • one year ago
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    was it multiple choice?

  32. anonymous
    • one year ago
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    with fill-ins

  33. myininaya
    • one year ago
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    oh ok cool stuff

  34. myininaya
    • one year ago
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    still trying to think what we can do for p>1

  35. myininaya
    • one year ago
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    I know you already done it but I might play with a bit more

  36. anonymous
    • one year ago
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    :) that's cool

  37. myininaya
    • one year ago
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    @jtvatsim do you have anything for p>1 ?

  38. jtvatsim
    • one year ago
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    Still working on it... Mathematica says to use a comparison test, but doesn't specify what to compare it to.

  39. jtvatsim
    • one year ago
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    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...

  40. anonymous
    • one year ago
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    i actually tried that, getting rid of the logs...but got stuck on that

  41. anonymous
    • one year ago
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    for limit comparison test, compare it to 1/k?

  42. jtvatsim
    • one year ago
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    tried that... lim comp fails... goes to infinity.

  43. jtvatsim
    • one year ago
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    Ah... might have something.

  44. jtvatsim
    • one year ago
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    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.

  45. myininaya
    • one year ago
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    are you saying: \[\frac{1}{(\ln(k))^e}=\frac{1}{k}?\]

  46. jtvatsim
    • one year ago
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    dang it... lol

  47. jtvatsim
    • one year ago
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    Assume (ln k)^e = 1. Then the result follows. LOL. That's called seeing what you want to. :)

  48. myininaya
    • one year ago
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    I totally get that. I do that sometimes.

  49. myininaya
    • one year ago
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    Then everyone else is like math is really magic.

  50. jtvatsim
    • one year ago
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    Yep, 0 = 1 implies anything.

  51. myininaya
    • one year ago
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    I'm kind of stumped

  52. jtvatsim
    • one year ago
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    Yeah, this is a tough question. There must be some trick involved here.

  53. anonymous
    • one year ago
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    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... ?

  54. jtvatsim
    • one year ago
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    Really? Cuz, that didn't work for me. The way my math is going right now though, I guess that may be true...

  55. jtvatsim
    • one year ago
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    Well, I am done for tonight. Have a good one everyone! :)

  56. anonymous
    • one year ago
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    goodnight, thanks for helping!~

  57. myininaya
    • one year ago
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    goodnight @jtvatsim

  58. myininaya
    • one year ago
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    @cstarxq sadly I guess I'm not picking the right series for comparison I give up for tonight :(

  59. myininaya
    • one year ago
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    oh i think i see I guess i forgot all the parts of the comparison limit test

  60. myininaya
    • one year ago
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    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

  61. myininaya
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx I was using pauls notes and did not see this for the limit comparison test.

  62. myininaya
    • one year ago
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    That was my guide until I did some more searching.

  63. myininaya
    • one year ago
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    and found that pdf

  64. anonymous
    • one year ago
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    @myininaya oh wow, that pdf will be really helpful for studying for my upcoming exam. Thanks for sharing!

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