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Homework Help: Series convergence, root test

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Does it converge, and what is the sum:

    [tex] \sum_{n=1}^{\infty}\frac{1}{n n^{\frac{1}{n}}} [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Please check my method and conclusion:

    Using the root test:

    [tex] \displaystyle\lim_{n\to\infty}\left|\frac{1}{n n^{\frac{1}{n}}} \right|^{\frac{1}{n}} = \displaystyle\lim_{n\to\infty}\frac{1}{n n^{\frac{1}{n}}} = \displaystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\lim_{n\to\infty}\frac{1}{ n^{\frac{1}{n}}} = 0 [/tex]


    so while this doe not give me the sum (I'm not excited about that part...), it at least tells me that this series converges. Am I correct?
     
  2. jcsd
  3. Feb 3, 2010 #2

    vela

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    Your first step is incorrect. How did you just get rid of the 1/n exponent?
     
  4. Feb 3, 2010 #3

    Mark44

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    I don't see that you actually took the root. You lost the 1/n exponent right away without changing anything.
    You could have simpified things by writing nn1/n as n1+1/n. This looks like a convergent p-series to me.
     
  5. Feb 3, 2010 #4

    Dick

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    NOT a convergent p-series, Mark44. The limit of n^(1/n) as n->infinity is one. Think about a comparison test.
     
  6. Feb 3, 2010 #5

    Mark44

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    I had a suspicion that it wasn't a p-series, given that the exponent wasn't a constant. Thanks for straightening me out, Dick.
     
  7. Feb 3, 2010 #6
    I'm not sure it matters that the exponent is a constant, as long as it is an integer n>0, because no matter what n is here, the value will be greater than 1. BUT since 1/n converges to 0, then we have the exponent converging to one, and then we have a problem.

    I'm still not sure what to do here. As far as comparison goes:

    [tex] \sum \frac{1}{nn^{\frac{1}{n}}} \leq \sum \frac{1}{n} [/tex]

    But that is not helpful b/c the 1/n series diverges...
     
  8. Feb 3, 2010 #7

    Dick

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    You've got (1/n)*(1/n^(1/n)). The second factor approaches 1. So for large n the second factor is greater than, say, 1/2.
     
  9. Feb 3, 2010 #8
    [tex] \sum \frac{1}{nn^{\frac{1}{n}}} \geq \sum\frac{1}{2n} [/tex]

    Since we know that the second series is divergent and smaller than the original, we can say the original must diverge as well.
     
  10. Feb 3, 2010 #9

    Dick

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    Best possible outcome, really. Now you don't have to worry about summing it.
     
  11. Feb 4, 2010 #10
    :-) Thanks for the input.
     
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