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Math Help - Integral Test

  1. #1
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    Integral Test

    I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
    I think there using the quotient rule but i cant seem to work it out.
    Can someone help me work out the problem?

     \sum_{n=1}^\infty  \frac 1  {\sqrt {n} ( \sqrt {n} + 1 )}

    then,


    Let  f(x) =   \frac 1  {\sqrt {x} ( \sqrt {x} + 1 )}

    The solution manual then shows!


     f'(x) =   \frac {1+2\sqrt x}  { 2x^{3/2} ( \sqrt x + 1 )^2} < 0

    With the denominator squared it does show they use the quotient rule. However, can you please
    show me the steps on how they derive the answer?
    Because i cant seem to get it. Its kinda a nasty problem.
    Thank you so very much!
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  2. #2
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    Re: Integral Test

    Quote Originally Posted by icelated View Post
    I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
    I think there using the quotient rule but i cant seem to work it out.
    Can someone help me work out the problem?
     \sum_{n=1}^\infty  \frac 1  {\sqrt {n} ( \sqrt {n} + 1 )}
    Note that \int_1^\infty  {\frac{{dx}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} is the same as \int_{2}^\infty  {\frac{{2du}}{u}} if u=\sqrt{x}+1.
    I have no idea what that solution manual could mean.
    Last edited by Plato; April 15th 2012 at 04:23 PM.
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  3. #3
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    Re: Integral Test

    Quote Originally Posted by icelated View Post
    I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
    I think there using the quotient rule but i cant seem to work it out.
    Can someone help me work out the problem?

     \sum_{n=1}^\infty  \frac 1  {\sqrt {n} ( \sqrt {n} + 1 )}

    then,


    Let  f(x) =   \frac 1  {\sqrt {x} ( \sqrt {x} + 1 )}

    The solution manual then shows!


     f'(x) =   \frac {1+2\sqrt x}  { 2x^{3/2} ( \sqrt x + 1 )^2} < 0

    With the denominator squared it does show they use the quotient rule. However, can you please
    show me the steps on how they derive the answer?
    Because i cant seem to get it. Its kinda a nasty problem.
    Thank you so very much!
    f(x) = \frac{1}{\sqrt{x}(\sqrt{x}+1)} = \frac{1}{x+\sqrt{x}} = (x + \sqrt{x})^{-1}

    chain rule ...

    f'(x) = -(x+\sqrt{x})^{-2} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)

    f'(x) = - \frac{1 + \frac{1}{2\sqrt{x}}}{(x+\sqrt{x})^2}

    f'(x) = - \frac{1 + \frac{1}{2\sqrt{x}}}{(x+\sqrt{x})^2} \cdot \frac{2\sqrt{x}}{2\sqrt{x}}

    f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}(x+\sqrt{x})^2}

    f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}[\sqrt{x}(\sqrt{x}+1)]^2}

    f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}(\sqrt{x})^2(\sqrt{x}+1)^2}

    f'(x) = - \frac{2\sqrt{x} + 1}{2x^{3/2}(\sqrt{x}+1)^2} < 0

    you left off the negative sign
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