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Convergence test: Factorials

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex]\sum \frac{n!}{10^n}[/tex]

    converges or diverges.


    (Note, I was unsure of how to format this via latex, so the summation is from n = 1 to infinity.)


    2. Relevant equations

    The root test:

    [tex]|\frac{a_n_+_1}{a_n}|[/tex]

    3. The attempt at a solution

    [tex] a_n=\frac{n!}{10^n},

    a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}[/tex]

    Applying the ratio test,

    [tex]|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1} [/tex]

    Cancelling terms out,

    [tex]\frac{n+1}{10} = r[/tex]

    Now, I know that if:

    r > 1, it is divergent,
    r < 1, convergent,
    r = 1, inconclusive.

    My problem is that I am not sure where to go after this. I still have an "n" in my answer, and I expected to just have a numerical answer.

    I was going to just go ahead and say that since n approaches infinity, r is greater than 1, and thus the series is divergent, but I stopped because I realized that from n = 1 to n = 8, the series would be convergent, and worse still, at n = 9, the test would be inconclusive?

    So my question is, did I make a mistake somewhere here, or is the ratio test not applicable here for this reason?

    Thanks for any feedback!
     
  2. jcsd
  3. Apr 4, 2010 #2

    Mark44

    Staff: Mentor

    It doesn't matter what happens between 1 and 8; you're interested in the behavior for large n. Every finite series converges, since you're just adding a finite number of terms.
     
  4. Apr 4, 2010 #3
    Alright, so I was right - it diverges.

    Thanks very much!
     
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