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Calculate lim n->inf (n^2/n!)

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    calculate lim n->inf (n^2/n!)

    2. Relevant equations



    3. The attempt at a solution
    i just calculated, and i got inf as the ans...
    n^2/1X2X3...Xn
    n/1X2X3...X(n-1)
    1/[(1X2X3...)/n](1-1/n)
    when n->inf, it will go to inf

    am i correct? or there is something wrong?
     
  2. jcsd
  3. Apr 20, 2015 #2

    SammyS

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    That is incorrect.
     
  4. Apr 20, 2015 #3
    what is wrong?
     
  5. Apr 20, 2015 #4

    Mark44

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    Your answer is wrong. The limit is not infinity.

    The 2nd and 3rd lines above are OK but written ambiguously.
    According to the standard rules of precedence, n/1X2X3...X(n-1) = ##\frac n 1 2 \cdot 3 \cdot \dots (n - 1)##, which is probably not what you intended.
    The 3rd line would be interpreted as ##\frac 1 { \frac{1 \cdot 2 \cdot 3 \dots } n} (1 - 1/n)##. That's so convoluted that you probably misled yourself.
     
  6. Apr 20, 2015 #5
    why not
    n / [1X2X3...(n-1)]
     
  7. Apr 20, 2015 #6

    Mark44

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    That would be fine, although at the calculus level, you should get out of the habit of writing X for multiplication. It's too easily confused as a variable.

    So what is ##\lim_{n \to \infty}\frac n {1 \cdot 2 \cdot 3 \cdot \dots (n - 1)}##?
     
  8. Apr 20, 2015 #7
    lim→∞ 1/ [ (1⋅2⋅3⋅⋅⋅/n)(1-1/n)] = 1/0 = inf ??
     
  9. Apr 20, 2015 #8

    Ray Vickson

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    You tell us.
     
  10. Apr 20, 2015 #9

    Mark44

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    @cloveryeah, here are some easier examples for you, to help you get your head around this stuff.

    ##\frac {3^2} {3!} = \frac 9 6 = 3/2##
    ##\frac {4^2} {4!} = \frac {16}{24} = 2/3##
    ##\frac {5^2} {5!} = ?##
    ##\frac {6^2} {6!} = ?##
    Do you see any sort of trend here?
     
  11. Apr 20, 2015 #10

    pasmith

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    Consider: If [itex]n > 4[/itex] then [itex]n! = n(n-1) \times \cdots \times 1 > n(n-1)[/itex]. Thus [itex]0 < \frac{n^2}{n!} < \frac{n^2}{n(n-1)} < \frac43[/itex]. That tells you that the limit of [itex]\frac{n^2}{n!}[/itex], if it exists, must be finite. Unfortunately this bound on [itex]\frac1{n!}[/itex] is not sufficiently accurate for the squeeze theorem to assist, since [itex]\lim_{n \to \infty} \frac{n}{n-1} = 1[/itex].

    But how, for [itex]n > 5[/itex], could you modify the above argument so that the squeeze theorem will assist?
     
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