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Homework Help: Converging function

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data

    determine whether [tex]\sum[/tex] n! / (n+2)! converges or diverges. if it converges, give the limit.


    3. The attempt at a solution

    i can see that this converges and the limit is 1/2

    but i dunno how to mathematically show it... how can i factorise out the 1/2? so i have 1/2!*n!/n! which would give 1/2!*1=1/2

    because n! / (n+2)! is not equal to n!/ (n!+2!)...
     
  2. jcsd
  3. Sep 25, 2010 #2
    Do you mean

    1. [tex]\sum\frac{n!}{(n+2)!}[/tex] or

    2. [tex]\frac{\sum n!}{(n+2)!}[/tex]

    ?

    If #1, can't you simplify [tex]\frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)}[/tex] ?
     
    Last edited: Sep 25, 2010
  4. Sep 25, 2010 #3
    the first one.

    and i dunno, can you do that? ive never done any work with factorials. i had to google what they were!
     
  5. Sep 25, 2010 #4

    Mark44

    Staff: Mentor

    Of course you can do that, based on how factorials are defined, usually something like this:
    0! = 1
    n! = n * (n - 1)!

    So (n + 2)! = (n + 2) * (n + 1)! = (n + 2)(n + 1)*n!
     
  6. Sep 25, 2010 #5
    " So (n + 2)! = (n + 2) * (n + 1)! = (n + 2)(n + 1)*n! "

    so then 1 / (n + 2)(n + 1) * n!/n!
    = 1 / (n+2)(n+1)

    = 1 / (n2 + 3n +2)
    now it looks like as n gets large, the limit is 0..
     
  7. Sep 25, 2010 #6

    Mark44

    Staff: Mentor

    Yes, but that tells you exactly nothing. If you have a series [tex]\sum_{i = 1}^\infty}a_n[/tex] and [tex]\lim_{n \to \infty}a_n = 0[/tex], the series could converge or it could diverge.

    Do you know any convergence tests you can use?
     
  8. Sep 25, 2010 #7
    ratio and comparison testing would only prove convergence, without giving the limit.

    integral test could work... let f(x) = 1/(n2+3n+2)

    [tex]\int[/tex] f(x).dx with upper bound infinity (let that be a letter, say b) and lower bound 1. then take the definite integral and evaluate the limit as b gets large.
     
  9. Sep 25, 2010 #8
    and that integral is = log(x+1) - log(x+2)
    = log[(x+1)/(x+2)]

    am i goin the right direction? i dunno where to go from there..
     
  10. Sep 26, 2010 #9

    Mark44

    Staff: Mentor

    The problem first asks you whether the series converges or diverges. Have you come up with a good reason for thinking that this series converges. "I can see that it converges" is not a good reason.

    The integral test tells you only that a series converges or diverges. It doesn't give you the sum of a convergent series.

    Hint: partial fractions...
     
  11. Sep 26, 2010 #10
    use the limit comparison test, with bn = 1/n2... this will show it converges.
     
  12. Sep 26, 2010 #11
    but then im still not sure how to evaluate the limit to show the limit of convergence.... if i multiply through by 1/n i end up with 0 as the numerator as n gets large, so that doesnt work. i dont know how else to go about it. bit dense when it comes to these.
     
  13. Sep 26, 2010 #12

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Follow through on Mark44's hint about partial fractions. Use partial fractions to write the partial sums as a difference between two terms. You will find some cancellations.
     
  14. Sep 26, 2010 #13
    like this...

    1/((n+1)(n+2))
    by decompostion becomes 1/(n+1) - 1/(n+2)

    so (1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....(1/(n+1)-1/(n+2))

    = 1/2 - 1/(n+2)

    as n gets large, 1/(n+2) becomes 0 so:

    = 1/2-0
    =1/2

    there.
     
  15. Sep 26, 2010 #14

    Mark44

    Staff: Mentor

    That's the basic idea, yes. Here's what the cleaned-up version looks like, using partial sums.

    [tex]S_n = \sum_{k = 1}^n \frac{1}{(k + 1)(k + 2)} = \sum_{k = 1}^n \left(\frac{1}{k + 1} - \frac{1}{k + 2} \right) = \frac{1}{2} - \frac{1}{n + 2}[/tex]
    [tex]\lim_{n \to \infty} S_n = \frac{1}{2}[/tex]
     
  16. Sep 26, 2010 #15
    yeahh im not the best at using the functions and stuff on here haha. thanks for the help though!
     
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