1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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


    User Avatar
    Gold Member

    Whenever factorial is involved in series, the ratio test is your best bet. :smile:
  7. May 7, 2012 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Convergent or Divergent?
  1. Divergent or Convergent? (Replies: 17)

  2. Converge or diverge (Replies: 7)