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Absolute convergence: ratio/root test n!/n^n

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Doing some problems from textbook, I need to determine whether the series is absolutely convergent, conditionally convergent, or divergent.

    n!/n^n

    I plugged it into WA, and it says the series doesn't converge, but I'm not sure how to figure it out.

    2. Relevant equations


    3. The attempt at a solution

    First, I applied the root test

    lim n->inf [tex]\frac{(n+1)!}{(n+1)^n} * \frac{n^n}{n!}[/tex]

    lim n->inf [tex]\frac{(n+1)n!}{(n+1)(n+1)^n} * \frac{n^n}{n!}[/tex]

    I reduce this, and apply the root test:

    lim n->inf [tex]\sqrt[n]{\frac{n^n}{(n+1)^n}}[/tex]

    lim n->inf [tex]\frac{n}{n+1}[/tex]

    lim n->inf [tex]\frac{1}{1 + 1/n}[/tex]

    = 1

    1 means that it's inconclusive. I'm not sure if I applied the tests incorrectly or if I'm supposed to try something else.
     
  2. jcsd
  3. Nov 24, 2009 #2
    It seems you took the root of the ratio. That's wrong. Don't combine the two tests. Use one or the other.

    The ratio test will work. (Your first step has a typo, but the second step has fixed it.) To finish it off, observe

    [tex]\frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n} \right)^n}=\frac{1}{\left( 1+\frac{1}{n} \right)^n}[/tex].

    The last expression has a famous limit.

    Actually the series converges, and as a double check using the comparison test, it is less than 2/n^2.
     
  4. Nov 24, 2009 #3
    Thanks for mentioning the typos, I see them. I meant to say I applied the ratio test first, and the exponent in the denominator of my first line should be n+1.

    I remember the limit from deriving it with L'Hopital's rule, 1/e. Thanks, I simply didn't see I could reduce it by dividing by n^n/n^n. :frown:
     
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