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Sequence - converges/diverges

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\infty\sum[/tex][tex]\frac{n^n}{(2n)!}[/tex]
    n=1

    First of all sorry for the bad attempt to replicate the problem digitally, but i hope you get the general idea. :)

    I just started learning sequences and i encountered this problem and im not exactly sure how to solve it.

    3. The attempt at a solution
    Here is the attempt:

    http://img38.imageshack.us/img38/215/captureco.jpg [Broken]http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]

    So i think i tried to use the "ratio test" (not sure how its called in english), to test if the limit is > or < than 1.

    When i got so far i have no clue what to do next, so i`m assuming i`m going in the wrong direction. Maybe the ratio test isn't the one to use here.

    Any tips ?


    Thanks in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 8, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The first two things in your "equation" above are not equal. I think you mean that the first is the sum itself and the next is the ratio. Please don't write "=" between things that are not equal!
    Since the general term is [itex]a_n= n n^n/(2n)![/itex], [itex]a_{n+1}/a_n= ((n+1)(n+1)^{n+1}/(2n+2)!)/(n n^n/(2n)!)[/itex][itex]= ((n+1)(n+1)^{n+1}/(2n)!)((2n)!/(n n^n)[/itex].

    Separate those as [itex][(n+1)/n ][(n+1)^{n+1}/n^n][(2n)!/(2n+2)!][/itex].

    (n+1)/n clearly goes to 1 and (2n)!/(2n+2)!= (2n)!/((2n)!(2n+1)(2n+1)= 1/(2n+1)(2n+2) goes to 0 "quadratically" so the real question is about whether [itex](n+1)^{n+1}/n^n[/itex] goes to infinity and, if so, how fast. [itex](n+1)^{n+1}= (n+1)^n(n+1)[/itex] so we are looking at [itex]\left((n+1)/n\right)^n (n+1)[/itex]. Does that go to infinity and, if so, how fast?

     
    Last edited by a moderator: May 4, 2017
  4. Nov 8, 2009 #3
    Thank you for the quick response, i didn't mean to write that the first two things were equal it just happened to be the easiest way to write. :)

    One question - why did you write that the general term is (n*n^n)/(2n)!, shouldn't it be (n^n)/(2n)! ?

    Well i thought of this way:
    We write this limit
    http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]
    as
    http://img63.imageshack.us/img63/899/capturesm.jpg [Broken]

    then we divide it like :

    http://img407.imageshack.us/img407/8123/capturepn.jpg [Broken] which is = 1/2

    and http://img18.imageshack.us/img18/323/capturelx.jpg [Broken] = http://img524.imageshack.us/img524/997/captureue.jpg [Broken] = goes to 1.

    (i didn't write the lim's in front.)

    So in the end 1 * 1/2 = 1/2 and 1/2 < 1 so => the sequence converges.

    Edit: Could someone confirm this ?
     
    Last edited by a moderator: May 4, 2017
  5. Nov 8, 2009 #4

    lanedance

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    Homework Helper

    hmmmm.. not sure if you missed it but as Halls pointed out:

    in your ratio, when:
    [tex] n \rightarrow n+1 [/tex]

    the the factorial becomes
    [tex] (2n)! \rightarrow (2(n+1))! = (2n+2)![/tex]
     
  6. Nov 8, 2009 #5
    [tex](n+1)^2=n^2+2n+1[/tex] but if the exponent is not 2, then the expansion has many more terms.


    Instead, write [tex]\frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n[/tex] and then this has a famous limit.
     
  7. Nov 8, 2009 #6
    Thanks for pointing that out, yes i missed that. But the good thing is that it doesn't affect the outcome. It`s still 1/2.

    And thanks Billy Bob.

    So that means:
    [tex]
    \frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n= \left(1+\frac{1}{n}\right)^n = e
    [/tex]

    Ahh and then it means that 1/2 * e = e/2 and e/2 > 1 and it means that it diverges. Though according to the book the answer is that it converges.

    Seems that i have missed something else.

    Damn this is frustrating.
     
  8. Nov 8, 2009 #7

    lanedance

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    Homework Helper

    have you included the extra n factor the revised factorial gives you into account in your ratio?

    if it really did go to C before, now it should be something like C/n...
     
  9. Nov 8, 2009 #8
    I was wrong after all.

    As lanedance pointed out i get [tex] (2n)! \rightarrow (2(n+1))! = (2n+2)![/tex] instead of [tex] (2n)! \rightarrow (2n+1)![/tex], and this is quite a huge deal.

    So in the end i get [tex] 1/(2n+2)(2n+1)[/tex] which is [tex](1/(2n+2))*(1/2n+1)[/tex] and both of these goes to 0.

    So i get 1/2 * e * 0 * 0 = 0 and 0 < 1 and that means that it converges.

    This forum is awesome, thanks guys. ;)
     
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