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Math Help - Integral Test

  1. #1
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    Integral Test

    infinity over Sigma n=1 1/n(n+1) Find whether it converges or diverges using the integral test. I know your suppose to use partial fractions but Im stuck on how to do that with an integral test.
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  2. #2
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    Quote Originally Posted by Link88 View Post
    infinity over Sigma n=1 1/n(n+1) Find whether it converges or diverges using the integral test. I know your suppose to use partial fractions but Im stuck on how to do that with an integral test.
    Fir check that the test applies (if a_n = f(n) is positive, continuous and decreasing)

    then consider

    \int _1^{\infty} f(x)\,dx

    In your case

    \int _1^{\infty} \frac{1}{x(x+1)}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x(x+1)}\,dx = \lim_{b \to \infty} \left. ln \left| \frac{x}{x+1} \right| \right|_1^b = \lim_{b \to \infty} ln \left| \frac{b}{b+1} \right| - \ln \frac{1}{2}
    = - \ln \frac{1}{2} = \ln 2

    which is finite so by the integral test the series converges.
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  3. #3
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    \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}<br />

    1 = A(x+1) + Bx

    x = 0 ... A = 1

    x = -1 ... B = -1

    \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}


    \lim_{b \to \infty} \int_1^b \frac{1}{x} - \frac{1}{x+1} \, dx<br />

    \lim_{b \to \infty} \left[\ln{x} - \ln(x+1)\right]_1^b<br />

    \lim_{b \to \infty} \left[\ln\left(\frac{x}{x+1}\right)\right]_1^b

    \lim_{b \to \infty} \left[\ln\left(\frac{b}{b+1}\right) - \ln\left(\frac{1}{2}\right)\right]

    \lim_{b \to \infty} \left[\ln\left(\frac{1}{1+\frac{1}{b}}\right) - \ln\left(\frac{1}{2}\right)\right] = 0 - \ln\left(\frac{1}{2}\right) = \ln(2)<br />
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  4. #4
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    Quote Originally Posted by Link88 View Post

    I know your suppose to use partial fractions
    \frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}, and that leads to danny's result.
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