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Limit as n->infinity of (n^n)/n!

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limit ##\lim_{n\to\infty} \dfrac{n^n}{n!}##

    2. Relevant equations
    Ratio test: ##a_{n+1}*\dfrac{1}{a_n}##
    3. The attempt at a solution
    I was having trouble evaluating this so I tried to use the ratio test which unfortunately leads to ##\lim_{n\to\infty} \dfrac{n}{n+1}## which just ends up being 1 which does not tell me whether the series is convergent or divergent. Alternatively I tried looking at it like this but I couldn't see where to go from there $$\lim_{n\to\infty} \frac{n}{1} \frac {n}{2} \frac {n}{3} \cdots \frac{n}{n} $$
     
  2. jcsd
  3. May 18, 2015 #2

    Orodruin

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    You can use either the last expression that you gave to bound the expression from below by a diverging function, or you can take the logarithm of everything and apply Stirling's approximation.
     
  4. May 18, 2015 #3
    Sorry I'm quite lost as to how I would go about finding a boundary for this function, is there a theorem or something I should be aware of?
     
  5. May 18, 2015 #4
    Hi potatochip911,
    n^n/n! = (n*n*....n)/(1*2*....n)
    (n+1)^(n+1)/(n+1)! = (n+1)(n+1)...../(1*2*...(n+1)),
    Hope this makes it easier, you can take out (n+1) from the second and compare the two series, and use induction, if the n^n/n! Is divergent then so is (n+1)^(n+1)/(n+1)!
     
  6. May 18, 2015 #5

    Orodruin

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    How about n/k > 1 for n > k?
     
  7. May 18, 2015 #6
    I don't think I'm allowed to use induction to solve these problems.

    Can I just do this? $$\frac{n^2}{2} < \frac{n^n}{n!} \mbox{on the interval} [1,\infty] \\ \lim_{n\to\infty} \frac{n^2}{2}=\infty$$ and then since a function smaller function diverges the original must also diverge? Edit: I think I messed this up again I keep thinking in terms of sums
     
  8. May 18, 2015 #7

    Orodruin

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    You can even do this: ##n = n/1 < n^n/n!##
     
  9. May 18, 2015 #8

    PeroK

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    You don't need induction, although it's perfectly valid to use induction for any maths problem - unless it's explicitly stated that you can't. Anyway, back to your problem. Where did you get ##n^2/2## from?

    ##n^n## and ##n!## both have ##n## terms; ##n^n## is clearly much larger than ##n!## as ##n## increases. So, all you need is a crude estimate. There is, in fact, a very simple estimate. Can you spot it?

    PS: I see Orodruin beat me to it!

    PPS: You are also allowed to calculate some values. For example, if you tried n = 5, you'd see that:

    ##5! = 120## whereas ##5^5 = 25 \times 125##

    So, it should be fairly obvious from that where this sequence is heading!
     
    Last edited: May 18, 2015
  10. May 18, 2015 #9

    Orodruin

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    I would say he used the first two factors of:
    rather than just the first.

    Edit: Of course, this does not tell you how fast it diverges. Stirling's approximation will do that for you (and you should find that it diverges exponentially).
     
  11. May 18, 2015 #10
    Okay I will keep that in mind for the future.

    Yea its just I'm going through a textbook from start to finish and we haven't touched on induction yet so I didn't want to use it. Thanks for the help guys!
     
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