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Determine If Integral Test Can Be Applied

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I attached the infinite series the question provided as a file.


    2. Relevant equations



    3. The attempt at a solution
    I deduced the general term to be ln(n)/n, so the infinite series would be written as [itex]\sum_{n=2}^{\infty} \frac{\ln(n)}{n}[/itex]

    I took the derivative of the general term, and I found that the function is not decreasing on the entire interval I am summing on. The answer key, however, says the derivative is negative on the entire interval. Could someone please help me?
     

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  2. jcsd
  3. Oct 6, 2012 #2

    SammyS

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    Well, what did you get for the derivative?
     
  4. Oct 6, 2012 #3
    (1-ln(x))/x^2
     
  5. Oct 6, 2012 #4

    SammyS

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    [itex]\displaystyle \frac{1-\ln(x)}{x^2}<0[/itex] for x > e .

    So evaluate, [itex]\displaystyle \sum_{n=2}^{\infty} \frac{\ln(n)}{n}=\frac{\ln(2)}{2}+\sum_{n=3}^{ \infty} \frac{\ln(n)}{n}\ ,[/itex] since 3 > e.
     
  6. Oct 6, 2012 #5
    Oh, so I can adjust the interval I am summing on?
     
  7. Oct 6, 2012 #6
    I have another one I am working on. (The solution from the book is attached as a file). If the function that is comparable to the general term of this series is decreasing for values
    x > e^1/2, why is one of the limits of integration x =1?
     

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  8. Oct 6, 2012 #7

    SammyS

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    So then, evaluate the integral from 2 to infinity.
     
  9. Oct 7, 2012 #8
    So, is it technically improper to evaluate the integral from 1 to infinity?
     
  10. Oct 7, 2012 #9

    SammyS

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    Quote the integral test, word for word.
     
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