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Convergence and divergence help

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine whether the following converges or diverges:

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

    with the sum going from n=1 to n=infinity.

    2. Relevant equations

    Comparison Test, Ratio Test.

    3. The attempt at a solution

    This should be do-able with the above two tests, since those are the only ones covered in the book up to this point. I think it diverges but I can't seem to get anywhere with either test. Any help appreciated!
     
  2. jcsd
  3. Dec 9, 2008 #2
    Re: Convergence/divergence

    I think you are going to want to use the ratio test... try setting it up this way...

    [tex]\lim_{n\rightarrow\infty} \frac{\frac{(n+1)^{n+1}}{((n+1)+1)^{((n+1)+1)}}}{\frac{n^n}{(n+1)^{(n+1)}}}[/tex]
     
  4. Dec 9, 2008 #3

    Dick

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    Re: Convergence/divergence

    You can try that. But it's going to give you a limit of 1. So the test is inconclusive. What you really need to do is find a way to use the famous limit (1+1/n)^n=e.
     
  5. Dec 9, 2008 #4

    gabbagabbahey

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    Re: Convergence/divergence

    The ratio test is inconclusive (check for yourself using randomlogic's setup); why don't you show us what you've tried for the comparison test?
     
  6. Dec 9, 2008 #5
    Re: Convergence/divergence

    Indeed it is inconclusive.
     
  7. Dec 9, 2008 #6

    Dick

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    Re: Convergence/divergence

    Reasonable suggestion, nonetheless.
     
  8. Dec 10, 2008 #7
    Re: Convergence/divergence

    Okay thanks, I think i got it:


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

    using the fact that (1+1/n)^n tends to e.

    Now [tex]\frac{1}{e+1}\sum{\frac{1}{n+1}}[/tex] diverges, therefore our series must diverge too :-)


    Does that seem okay?
     
  9. Dec 10, 2008 #8

    Dick

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    Re: Convergence/divergence

    That's it. (1+1/n)^n is actually <=e, isn't it? But there's nothing wrong with playing it safe and saying <=(e+1) either.
     
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