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For the series

  1. Dec 17, 2007 #1
    (-1)^n/sqrt(n+2) , How do I show that it is conditionally convergent?

    The absolute value of the series, 1/n^(1/2) diverges, so shouldn't the series diverge?


    --Also, I have never encountered a problem like this, so here is another question:

    n(1+n^2)^p , for which values of p does the series converge?
    I tried the integral telst by letting u = 1 + n^2, but could not get the answer...
     
  2. jcsd
  3. Dec 17, 2007 #2

    HallsofIvy

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    Do you understand what "conditionally convergent" means? By showing that The series with absolute values diverges you have shown that the series is NOT "absolutely convergent". But a series with alternating sign is a whole different matter: use the "alternating series test": if a series is of the form (-1)nan and {an} is a decreasing sequence, then the series converges. A series that converges but does NOT converge absolutely is called "conditionally convergent".


    ??Do you mean a fraction with that in the denominator? If so then go ahead and multiply it out: you get a polynomial with leading term n2p+ 1. It's not too difficult to show that no matter what the other terms are, eventually that fraction is smaller than 2/n2p+1 and that converges, absolutely for 2p+1> 1 or p> 0.
    If you actually do mean the series n(1+ n2)p then it obviously does not converge for p> 0. If p is negative, it can be written in the form n/(1+ n2)-p. Now the denominator will have leading coefficient n2p and, dividing both numerator and denominator by n, is of the form 1/(n2p-1+ ...). That is clearly larger than 2/n2p-1 and so converges absolutely for 2p-1> 1 or p> 2.
     
  4. Dec 17, 2007 #3
    The answer should be p < -1...
    That is the series given (n(1+ n^2)^p ), there is no fractions involved.Thank you for the clarification on the first question, now the only thing I still don't get is how to do the second question. I tried using the ratio test, but it didn't get anywhere.
     
  5. Dec 17, 2007 #4

    HallsofIvy

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    Sorry, I had my signs wrong! n/(1+n^2)-p is, when you multiply that out, the same as n/(n(-2p+ other terms) which is the same as 1/(n-2p-1+ other terms). You can "compare" that to 1/n-2p-1 which, by the "p test" converges absolutely for -2p-1> 1 or -2p< 2 or p< -1.
     
  6. Dec 17, 2007 #5
    Thanks, but when I multiply what out?
     
  7. Dec 17, 2007 #6

    HallsofIvy

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    [itex](1+ x^2)^{-p}[/itex], of course! (Remembering that, in this case, -p is positive.)
     
  8. Dec 17, 2007 #7
    How do I multiply that out? I appreciate your patience but it is just not clear to me...
     
  9. Dec 18, 2007 #8

    HallsofIvy

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    How would you multiply [itex](1+ x^2)^2[/itex]? [itex](1+ x^2)[/itex]?

    Notice that the only thing you need to now its the leading term- the one with highest exponent.
     
  10. Dec 18, 2007 #9
    for that multiplication, [itex](1+ x^2)^2[/itex], shouldn't the highest exponent be x^4, why do you get n^(-2p-1) for the series, where does the -1 come from?

    Oh, I see you divided it by n, so it fits the form 1/n...
     
    Last edited: Dec 18, 2007
  11. Dec 18, 2007 #10
    You flipped the > sign too early in post 4.
     
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