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Sequence (n!)/(n^n) Convergent or Divergent and Limit?

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Is the sequence {(n!)/(n^n)} convergent or divergent. If it is convergent, find its limit.

    2. Relevant equations
    Usually with sequences, you just take the limit and if the limit isn't infinity, it converges... That doesn't really work here. I know I'm supposed to write out the terms, but I can't figure out where to go from there.

    3. The attempt at a solution
    (n!)/(n^n) = ((1)(2)(3)...(n-1)(n))/((n)(n)(n)...(n)(n))
    I think I'm supposed to say that this is less than something that goes to zero, but I don't know what that something should be.

    n/n = 1

    Does this mean that I can cancel that out and have a list of numbers smaller than n divided by n? Then the sequence is less than 1...but I think it converges to zero, I just don't know how to show that.
     
  2. jcsd
  3. Oct 29, 2014 #2

    RUber

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    You are on the right track. Try writing out a few terms to see if you can see the growth factor from ##\frac{n!}{n^n}## to ##\frac{(n+1)!}{(n+1)^{(n+1)}}##.
     
  4. Oct 29, 2014 #3
    Okay, so...

    ((n+1)!)/((n+1)^(n+1)) = (1/(n+1)) ((n+1)!)/((n+1)^n) = (1/(n+1)) (((1)(2)(3)...(n)(n+1))/((n+1)(n+1)(n+1)...(n+1)(n+1)) < (1/(n+1)) and the limit of 1/(n+1) as n goes to infinity is zero...

    Is my inequality true? My logic was just that the huge set of terms being multiplied by each other is still less than 1, so a (1/(n+1)) is going to be greater than itself multiplied by a number less than one. Is this correct?

    How exactly does this relate to n!/(n^n)?
     
  5. Oct 30, 2014 #4

    haruspex

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    RUber is suggesting you consider the ratio between consecutive terms, [itex] \frac {n!}{n^n} [/itex] and [itex]\frac {(n+1)!}{(n+1)^{n+1}}[/itex].
     
  6. Oct 30, 2014 #5

    Zondrina

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    The ratio test would work I believe. In general though: ##n^n > n! > b^n > n^x > \ln(n)## as ##n \rightarrow \infty##. So its easy to determine convergence by merely examining the behaviour of ##a_n##.
     
  7. Oct 30, 2014 #6

    RUber

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    I think you have said it for the n+1 case. The logic holds for the nth case just the same.
    ##\frac{n!}{n^n}=\frac1n \frac2n ... \frac nn ## the largest term in the product is 1, so the product can be no larger than the smallest term. Just as you said in post #3.
     
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