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Check my work for.

  1. Mar 27, 2009 #1
    For which positive integers k is the following series convergent?
    [tex]\sum(n!)^2/(kn)![/tex]


    attempt: use the ratio test.

    let Bn = (n!)2 / (kn!)

    let Bn+1 = ((n+1)!)^2 / (k(n+1))!


    then Bn+1 / Bn =

    ((n+1)!)^2 / (k(n+1))! * (kn)!/(n!)^2


    from manipulation I got :

    (n+1)2 / k(n+1)

    = (n+1) / k


    if this is right (can you check please) the how would it follow to determine K such that
    the following series is convergent?
     
    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2

    Mark44

    Staff: Mentor

    Your answer is right, but needs parentheses: (n + 1)/k. As you wrote it, it would be interpreted as n + (1/k), and I don't think that's what you meant.

    For the ratio test to show convergence, it must be true that
    [tex]\lim_{n \rightarrow \infty} \frac{n + 1}{k} < 1[/tex]

    Is that going to happen here?
     
  4. Mar 28, 2009 #3
    I guess not. Because for any number k, if you take the limit, then the limit
    of the numerator is going to be infinity, and the denominator is just going to make it still
    infinity. How would you prove this deductively?
     
  5. Mar 28, 2009 #4

    Mark44

    Staff: Mentor

    You've pretty much said what needs to be said. The idea is that k is some fixed number, so eventually, no matter how big k is, n will surpass it, and your ratio will be larger than 1.
     
  6. Mar 28, 2009 #5
    Wait, my online hw only gives the option

    k ? _____

    the ? is >= , <= .
     
  7. Mar 28, 2009 #6
    I don't think my work was correct. let me try again.

    Root test:

    [tex]\frac{((n+1)!)^2}{(k(n+1)!)^2}[/tex] x [tex]\frac{(kn)!}{(n!)^2}[/tex]

    =

    [tex]\frac{(n+1)^2 * (n!)^2}{k(n+1) * (kn)!}[/tex] x [tex]\frac{(kn)!}{(n!)^2}[/tex]

    =

    the kn! cancels out and the and the (n!)^2 cancels out

    and we have



    [tex]\frac{(n+1)^2}{k(n+1)}[/tex] < 1

    (n+1)2 < k(n+1)

    (n+1) < k

    same as before.

    maybe we should take a different approach.

    just from induction, how about k being (n+1), that would make the bottom denominator
    greater than the top.
     
  8. Mar 28, 2009 #7
    Just guessed and the answer is 2. Can you make out why, and not just from plugging in numbers
     
    Last edited: Mar 28, 2009
  9. Mar 29, 2009 #8

    Mark44

    Staff: Mentor

    I think you might not have given us all the information.

    Is this the series?
    [tex]\sum_{k = 0}^\infty \frac{(n!)^2}{(kn)!}[/tex]

    You didn't include the index in your post, and I was assuming the index was n. If it's k, that's a different story.
     
  10. Mar 29, 2009 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor


    Caution: (kn!) is not the same as (kn)!

     
  11. Mar 29, 2009 #10
    Nope the index is n=1 to infinity. And for HallsofIvy, I meant to write (kn)! and not kn!.

    Well if I look at the series I can see why the answer is 2 because the denominator will also be n^2.
     
  12. Mar 31, 2009 #11
    I think this move is invalid :

    [tex]
    \frac{((n+1)!)^2}{(k(n+1)!)^2}[/tex]

    =>

    [tex]
    \frac{(n+1)^2 * (n!)^2}{k(n+1) * (kn)!}[/tex]

    changing the (k(n+1))! to
    k(n+1) * (kn)!

    This has to invalid, otherwise it does not makes any sense.
     
  13. Apr 6, 2009 #12
    no takers ?
     
  14. Apr 8, 2009 #13
    still want help if you know how to
     
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