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Integral test for converging series

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data
    In one of my problems, i'm asked to use integral test in order to determine whether the following function is converging or diverging:
    (summation from n=1 to inf) 1/(n^2-4n-5)
    I was just wondering how do we know if function is positive, continuous and decreasing because when i tried to see if it was positive (i plugged in a value), i got a negative...
    Also, for a integral test, does the entire function have to be decreasing?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 12, 2009 #2

    gabbagabbahey

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    Re: Series

    Is the summation really form n=1 to infinity? If so, I can tell you without testing that the series diverges (take a look at the the n=5 term:wink:)

    Anyways, for the integral test to be a valid test, your summand (function) must be monotone decreasing and non-negative on the entire summation interval. If you plug in a value of 'n' (on the summation interval) and get a negative number you should know right there that the integral test is useless.
     
  4. May 12, 2009 #3
    Re: Series

    You said that's it's apparently diverging, but in the example, we found it to be converging:(-1/6)ln(-2)...i dunno 'cause you mentioned that it has to be non-negative on its entire summation interval...:confused:
     
  5. May 12, 2009 #4

    gabbagabbahey

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    Re: Series

    There is no way this series converges for n=1 to n=infinty.... the problem is the n=5 term...

    [tex]f(5)=1/(5^2-4(5)-5)=1/0=\infty[/tex]

    How can a sum converge when one of its terms is undefined?!

    Also, [tex]\int_1^\infty 1/(x^2-4x-5)dx\neq\frac{-1}{6}\ln(-2)[/tex] ...the integral doesn't even converge!
     
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