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Series Comparison Test

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Find if the sum from n = 1 to infinity of (n^n)/n! diverges or not.

    2. Relevant equations

    p = an+1/an


    3. The attempt at a solution

    Using the comparison test I get to the point where p_n = (n+1)^(n+1) / [(n+1) n^n]

    Shouldnt p just be 0, dont (n+1)^(n+1) and n^n cancel for large n? The book says the answer is p = e, how do you get there?

    Thanks for the help!
     
  2. jcsd
  3. Oct 21, 2009 #2
    Why not try the ratio test?
     
  4. Oct 21, 2009 #3

    Office_Shredder

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    nn+1/nn doesn't cancel for large n, so when you replace nn+1 with (n+1)n+1 why would you expect it to?
     
  5. Oct 21, 2009 #4

    Dick

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    No, they don't cancel for large n. You are jumping to conclusions. If I look at your expression I would write it as (n+1)^n/n^n. Do you see how? Now what?
     
  6. Oct 21, 2009 #5

    lanedance

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    not zero, cancel the n+1 term
    [tex] \frac{(n+1)^{n+1}}{(n+1) n^n} = \frac{(n+1)^n}{n^n} = (\frac{n+1}{n})^n [/tex]
     
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