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Testing a sequence for converge or divergence

  1. Jul 22, 2011 #1
    I have 4 problems left and the questions says I have to test them for converge or divergence.
    Here are the problems
    http://gyazo.com/f0fa5a38c5968ecb7e74103486a181bd.png
    http://gyazo.com/4689f9d02d0b264c2c2b64ff4907ba77.png
    for 25, I want to take the limit as n goes to infinity however I get (-1)^[itex]\infty[/itex] and stuck there

    for 27, similar problem how do i take the limit of cos([itex]\infty[/itex])

    for 43 i have no idea

    for 45, I feel like i should use lhr, but i cant take the derivative of n!

    Please try and explain it to me, as I already have the answers I just want to learn
     
  2. jcsd
  3. Jul 22, 2011 #2
    25 - Have you done absolute convergence?
    26 - Do you see what the graph of this function looks like?
    45- Maybe you could try expanding out and then comparing it with some other sequence?
     
  4. Jul 22, 2011 #3

    Dick

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    For 25, I think the question is about convergence of the sequence, not convergence of the series. Ignore the (-1)^n for the moment and tell us whether the n/(n^2+1) part converges and to what.
     
  5. Jul 22, 2011 #4

    SammyS

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    What methods do you know? What does it mean for a sequence to converge? ... diverge?

    How did you get (-1) for 25 ? Show us your work, so we can help you.

    Write out the first several terms for 27.

    For 45, use logarithms.
     
  6. Jul 22, 2011 #5
    Oops my bad. Will think before I post next time hahah. =D
     
  7. Jul 22, 2011 #6
    yes, but this is a sequence I thought you cant use tests on it...?
    27- i looked at the graph and it seems its going to 0 but is a graph an accurate way of knowing
    45- i feel like theres a better way


    thats going to 0

    I think i know almost all the methods, but these are sequences so i thought you really cant use them..
    A sequence to converge means as it expands it eventually reaches a limit of a rational number, for it diverges means it goes to infinity or negative infinty

    for 25 well i got the limit as n goes to infinity of that function, but if i try to take the limit it becomes
    (-1)^[itex]\infty[/itex] *[itex]\infty[/itex]/[itex]\infty[/itex]

    27 - seems like its going to 0

    45- I would have the ln(n!) - ln(2^n) thats infinity - infinity, how can I rewrite that?
     
  8. Jul 22, 2011 #7

    Dick

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    This would be a lot less confusing if you work on the problems one at a time. I'd suggest you finish 25 first. -1<=(-1)^n<=1. Try using a squeeze argument. And I don't think the best answer for limit n/(n^2+1) is infinity/infinity. In fact, that's no answer at all. Didn't you also say it was 0? Why?
     
    Last edited: Jul 22, 2011
  9. Jul 22, 2011 #8

    SammyS

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    No, diverges simply means that it doesn't converge.
     
  10. Jul 22, 2011 #9

    SammyS

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    Use the properties of logarithms to simplify ln(n!) - ln(2n) .
     
    Last edited: Jul 22, 2011
  11. Jul 23, 2011 #10
    I dont see how that would help? can you show me the steps? wouldnt it just be ln(n!) - n*ln(2), which would still be infinity - infinity?
     
  12. Jul 23, 2011 #11

    Dick

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    It's probably easier and more direct to write n!/2^n as (1/2)*(2/2)*(3/2)*(4/2)*...*(n/2). Now notice, say, that there are lots of factors that are greater than 2 if n is large. Think about a comparison test.
     
  13. Jul 23, 2011 #12

    SammyS

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    Dick's idea is probably better, but here's what I had in mind.

    n! = n(n-1)(n-2)...(4)(3)(2)(1).

    Therefore, ln(n!) = ln(n)+ln(n-1)+ln(n-2)+...+ln(4)+ln(3)+ln(2) .

    But then comparing this to (n)ln(2) can be a bit messy.
     
  14. Jul 23, 2011 #13
    Wouldn't that be enough to show that the terms clearly diverge, or at least do not converge to 0?

    I myself would probably go with using the ratio test.
     
  15. Jul 23, 2011 #14

    SammyS

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    I believe these are sequences, not series, although the only place it says that is in the title.

    However, doing #45 via the logarithm does change it to being equivalent evaluating the convergence of an infinite sequence.
    ... which gives me an idea

    Group a -ln(2) with each ln(i) to give: ln(an) = (ln(n)-ln(2)) + (ln(n-1)-ln(2)) + (ln(n-2)-ln(2)) +...+ ln(4)-ln(2)) + (ln(3)-ln(2)) + (ln(2)-ln(2)) + (ln(1)-ln(2)) .
     
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