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Homework Help: Convergent or Divergent?

  1. May 7, 2012 #1
    I am to find whether the sum of (n!)/(n^n) converges or diverges. I tried both the limit comparison test, and a regular comparison test. (These are the only types of tests I am allowed to use.) So I tried several approaches:

    Approach #1: (n!)/(n^n) > 1/(n^n)

    Normally we use a setup like this to prove something with a p-series. However, the expression on the left side of the inequality isn't a p-series.

    Approach #2: (n!)/(n^n) < n!

    While this expression is true, it is not useful because the formula for the series is less than, not greater than, the series that is known to diverge.

    Approach #3 (Limit comparison): an = (n!)/(n^n) ; bn = 1/(n^n)

    an/bn = n!

    The limit here is, unfortunately, infinite, and I have to stop here.

    What other approach can I take that would result in more success?
  2. jcsd
  3. May 7, 2012 #2


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    Science Advisor

    Hey joe_cool2 and welcome to the forums.

    My intuitive guess is that your function will converge. One test to show this that seems appropriate would be the ratio test. Take a look at this page:

  4. May 7, 2012 #3
    Hello, thanks for the welcome. It is much appreciated.

    I am perfectly aware that it is often wise to use the ratio test in these situations with n!. However, I have been restrained to using specifically those two techniques mentioned earlier. Any ideas?
  5. May 7, 2012 #4
    The idea is to compare your function to a geometric series. That is, you have to find an [itex]0\leq a<1[/itex] such that

    [tex]\frac{n!}{n^n}\leq a^n[/tex]

    for all n. If you can find such an a, then your series will converge. So we need to find an a such that


    Try to show that the left hand side is decreasing from a certain point on.
  6. May 7, 2012 #5


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    Gold Member

    Whenever factorial is involved in series, the ratio test is your best bet. :smile:
  7. May 7, 2012 #6
    He's not allowed to use it. Please read the thread first before responding.
  8. May 7, 2012 #7
    Can you show that [itex]\sum\frac{n^n}{(n!)^n}[/itex] converges? If you can, then use the limit comparison test with that and your series.
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