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Limit of a sequence

  1. Sep 18, 2007 #1
    Problem: Find the limit of the sequence

    [tex]a_{n}=\frac{n^{2}2^{n}}{n!}[/tex]

    My first thought was to say that

    [tex]0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!} [/tex]

    and by squeeze theorem, since [tex]\frac{x^n}{n!}=0[/tex] for all real x, my original limit must be 0 as well. Now all I need to do is prove that [tex]n^{2}2^{n} \leq x^n[/tex], which is where I'm stuck. Can anyone give me a hand? Thanks.
     
    Last edited: Sep 18, 2007
  2. jcsd
  3. Sep 18, 2007 #2

    EnumaElish

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    Rearrange to n^2 < (x/2)^n.
     
  4. Sep 18, 2007 #3

    dynamicsolo

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    What is x? Is [tex]0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!} [/tex] even true?

    You might break up the numerator and denominator into their respective factors and look at the ratios that are formed. I believe you'll find that some have specific values, others have finite values in the limit n-> inf. , and others go to zero in the limit.
     
  5. Sep 18, 2007 #4

    EnumaElish

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    Not for any x and any n; but I think n^2 2^n < x^n as n --> +infinity.
     
  6. Sep 18, 2007 #5
    Sorry if it seems stupid, but I'm horrible at proofs, and am still coming up a bit short on proving [tex]n^2 \leq {(\frac{x}{2})}^n[/tex]

    I mean it seems fairly obvious that this is true to me, and more obvious still when I begin writing it out term by term for increasing n, but is this good enough to state it conclusively?
     
    Last edited: Sep 18, 2007
  7. Sep 18, 2007 #6

    Dick

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    dynamicsolo's advice is a bit better. Write it as n*n*4*2^(n-2)/(n*(n-1)*(n-2)!). Does that suggest anything?
     
  8. Sep 18, 2007 #7
    I must be missing something obvious, because I've tried that as well with pretty much no success. :frown:
     
  9. Sep 18, 2007 #8

    dynamicsolo

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    I'm suggesting that you write each factor in the numerator over a factor in the denominator and then consider each ratio you've formed. An example:

    (e^n)/(n!) = (e·e·...·e)/[n·(n-1)·...·3·2·1] ;

    there are n factors each in the numerator and denominator, so you can also write this as

    (e/n)·(e/n-1)·...·(e/3)·(e/2)·(e/1).

    In the case of our ratio, there are (n+2) factors in the numerator and only n in the denominator, so we will need to set aside two factors upstairs -- two of the 2s may be best. Now look at your ratio as a product of factors and recall that the limit of a product of terms is the product of the limits of the terms.

    Remember also that since we want the limit of the sequence a-sub-n, we are looking for the limit of this product as n-> infinity.
     
  10. Sep 18, 2007 #9

    Dick

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    What are the limits of n/n, n/(n-1) and coup de grace 2^(n-2)/(n-2)!, I thought you said you knew x^n/n! -> 0. x=2 and if n->infinity, n-2->infinity.
     
  11. Sep 18, 2007 #10
    I got it. Thank you all, you've been a big help.
     
  12. Sep 18, 2007 #11

    dynamicsolo

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    *thuds head with heel of hand, causing skull to emit dull, hollow sound*

    I just thought of something else: what is

    lim n->inf. of abs([tex]\frac{a_{n+1}}{a_{n}}[/tex])?

    We know the terms of this sequence are positive and you'll find this limit tends to zero. Thus the limit of the sequence is zero.
     
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