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Converge or diverge?

  1. Dec 9, 2007 #1
    Determine whether the following converge or diverge (show all work and state all tests and theormes used!)

    a)[tex]\infty[/tex] (n=1)[tex]\sum[/tex]cos[tex]\hat{}[/tex]2 n/n [tex]\sqrt{}[/tex]n

    b)[tex]\infty[/tex] (n=1)[tex]\sum[/tex]4(-1)[tex]\hat{}[/tex]n-1/n[tex]\hat{}[/tex]5+n[tex]\hat{}[/tex]3

    i need help answering these 2 problems!!! any help wld be appreciated! Thnx!!!
  2. jcsd
  3. Dec 9, 2007 #2


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    Staff Emeritus
    Science Advisor

    Firstly, you need to show your work; we're not here to do your homework for you! Secondly, your expressions are not clear. Here, I'll rewrite them for you; click on the images to see the code:

    [tex]\sum_{n=1}^{\infty}\frac{\cos^2n}{n\sqrt n}[/tex]

    [tex]\sum_{n=1}^{\infty}\frac{4(-1)^{n-1}}{n^5+n^3} [/tex]

    Are they the questions you are asking?

    Finally, please be sure to post any further questions in the homework help forums.

    Welcome to PF, BTW.
  4. Dec 9, 2007 #3
    thank you! yes those are the questions! it took me forever just to get them looking like that! Thank you again i'm still trying to get used to this forum....
  5. Dec 9, 2007 #4
    Hi, i suposse i would procceed in the following way:
    Note that due to the fact that cos(x) is always smaller or equal than 1,
    cos^2(x) has the same behavior. So, the sum that you are asking:
    [tex]\sum_{n=1}^{\infty}\frac{\cos^2n}{n\sqrt n}[/tex] is smaller than
    [tex]\sum_{n=1}^{\infty}\frac{\1}{n\sqrt n}[/tex] and this last converges
    because it's a "p-type" sum with p=3/2>1. So by comparission, the first sum
    must converge.
    In the second one, you may see that it's an alternant series (the 4 doesn't matter)
    and also that the sequence formed by the terms is a positive and decreasing one, so
    due to the Leibniz rule (that applies only in this case) the sum is convergent.

    Sory for the english i still don't know it very well
  6. Dec 9, 2007 #5
    there was supposed to be a 1 in the numerator of the second sum...
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