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Homework Help: Determine whethere the following series is convergent or divergent

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following series is convergent or divergent:
    [tex]
    \frac{1}{2^2}+\frac{2^2}{3^3}+\frac{3^3}{4^4}+....
    [/tex]
    I rewrite it as:
    [tex]
    \sum_{n=1}^{\infty} \frac{n^n}{(n+1)^{n+1}}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution
    I stopped.
    I can not do anything.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2010 #2

    jgens

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    Gold Member

    You do need to show us your attempt at the solution before we can help, so which convergence tests have you tried?
     
  4. Jan 24, 2010 #3
    Its not geometric
    not telescoping
    nth term test failed
    can not use integral test
    Ratio & Root failed because its algebric function

    There is only the comparison tests
    I tried them a lot
    but every time i failed :/
     
  5. Jan 24, 2010 #4

    Dick

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    Science Advisor
    Homework Helper

    Write it as (n^n/(n+1)^n)*(1/(n+1)). The first factor approaches a finite limit. What is it?
     
  6. Jan 24, 2010 #5
    The first one goes to 1/e
    the second goes to 0
    1/e times 0 = 0


    nth term test failed.
     
  7. Jan 24, 2010 #6

    jgens

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    Gold Member

    Alright, I would suggest using the comparison test for this particular series (although I'm sure that there's probably a simpler way to do it). If [itex]n \geq 1[/itex] can you find an upper bound for the following:

    [tex]\frac{(n + 1)^{n+1}}{n^{n+1}}[/tex]

    Edit: Fixed typo

    Edit 2: Fixed another typo. Sorry, I just seem to be mistake prone today.
     
    Last edited: Jan 24, 2010
  8. Jan 24, 2010 #7

    Dick

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    What 'nth term test' are you talking about? You are right the first factor approaches 1/e. So for large values of n a term in the series is approximately 1/(e*(n+1)). What kind of test does that suggest?
     
  9. Jan 24, 2010 #8
    nth terms test = Test for divergence.

    hmmm .. I got it .. Limit comparison test with the divergent series [tex]\sum_{n=1}^{\infty} \frac{1}{n+1}[/tex]
    The limit of the limit comparison test will give 1/e which is positive and finite number
    hence, our series diverges.

    Right?
     
    Last edited: Jan 24, 2010
  10. Jan 24, 2010 #9

    Dick

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    Science Advisor
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    Right.
     
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