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Factorial Limit

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data

    So, I'm doing a series problem, and after applying the root test I end up with the following limit:

    Lim as n--> inf of (n! / n^2) = infinity according to the solution manual.

    I can't seem to understand this because the way I look at it, each term in the numerator is getting smaller [ (n)(n-1)(n-2)..] but each term in the denominator is staying constant as n^2. So shouldn't the limit equal zero?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. May 22, 2015 #2

    Ray Vickson

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    Google 'properties of factorial function' for example. You will see exactly what is going on for large n.
     
  4. May 22, 2015 #3

    pasmith

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    For [itex]n \geq 3[/itex] we have
    [tex]
    \frac{n!}{n^2} \geq \frac{n(n-1)(n-2)}{n^2} = n - 3 + \frac{2}{n}.
    [/tex] The inequality follows from the fact that if [itex]n = 3[/itex] then [itex]n! = n(n-1)(n-2)[/itex] whilst if [itex]n \geq 4[/itex] then [itex](n-3) \times \dots \times 1 \geq 1[/itex].
     
  5. May 22, 2015 #4
    I recommand stirling approximation, it will be very appropriate for this case, it states that when n goes to infinity n! = √(2πn) * (n/e)^n, so try it !!
     
  6. May 22, 2015 #5

    vela

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    You're looking at the wrong end in the numerator. Look at the n=3, n=4, and n=5 terms, for example. You have
    \begin{align*}
    \frac{3!}{3^2} &= \frac{3\cdot 2\cdot 1}{3\cdot 3} \\
    \frac{4!}{4^2} &= \frac{4 \cdot 3\cdot 2\cdot 1}{4 \cdot 4} \\
    \frac{5!}{5^2} &= \frac{5\cdot 4 \cdot 3\cdot 2\cdot 1}{5 \cdot 5}
    \end{align*} Can you see the mistake in your reasoning?
     
    Last edited: May 24, 2015
  7. May 25, 2015 #6
    So let's say I have the sequence n!(2^n) /(2n)!, how would I show that this converges to zero.
     
  8. May 25, 2015 #7

    Mark44

    Staff: Mentor

    Start by expanding the (2n)! in the denominator to (2n)(2n -1) ...(n+1)(n!).
     
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