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Does the series with terms n!/e^n converge or diverge

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Does this series converge or diverge? infinity[tex]\Sigma[/tex]n=1 (n!/e^n)


    2. Relevant equations

    The ratio test.

    3. The attempt at a solution
    lim n--> infinity ((e^n)(n+1)!)/(e^(n+1))

    I don't know what to do from here...
     
  2. jcsd
  3. Feb 23, 2010 #2

    Mark44

    Staff: Mentor

    Re: Series

    You're missing the n!. Set up the ratio of a_(n+1)/a_n, simplify it, then take the limit.
     
  4. Feb 23, 2010 #3
    Re: Series

    This is what I have:

    lim n-->infinity [(n+1)!/(e^(n+1))][(e^n)(n!)]
    lim n-->infinity [(n+1)/(e)]

    Did I simplify right? And do I just plug in infinity now?
     
  5. Feb 23, 2010 #4

    Mark44

    Staff: Mentor

    Re: Series

    Yes, that's simplified correctly. No, you don't actually plug in infinity, but as n gets larger and larger, what happens to (n + 1)/e?
     
  6. Feb 23, 2010 #5
    Re: Series

    It also gets larger. Which means that the series diverges.
     
  7. Feb 23, 2010 #6

    Mark44

    Staff: Mentor

    Re: Series

    Yes and yes. Using the ratio test, you found that lim a_(n + 1)/a_n is infinity, and so the ratio test tells us this series diverges.
     
  8. Feb 23, 2010 #7
    Re: Series

    Yay! Thank you!
     
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