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Math Help - Integral Test

  1. #1
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    Integral Test

    Summation, n=1 to infinity (ne^(-n/2))

    Please help.
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  2. #2
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    Hello, saiyanmx89!

    Integral test: . \sum^{\infty}_{n=1} ne^{\text{-}\frac{n}{2}}

    The corresponding integral is: . \int^{\infty}_1 xe^{\text{-}\frac{1}{2}x}dx


    By parts: . \begin{array}{ccccccc}u &=& x & & dv &=& e^{\text{-}\frac{1}{2}x}dx \\ du &=& dx && v &=& -2e^{\text{-}\frac{1}{2}x} \end{array}

    We have: . -2xe^{\text{-}\frac{1}{2}x} + 2\int e^{\text{-}\frac{1}{2}x}dx \;=\;-2xe^{\text{-}\frac{1}{2}x} - 4e^{\text{-}\frac{1}{2}x} \;=\;\frac{-2(x+2)}{e^{\frac{x}{2}}}\,\bigg]^{\infty}_1

    Then: . \lim_{b\to\infty}\frac{-2(x+2)}{e^{\frac{x}{2}}}\,\bigg]^b_1 \;=\;\lim_{b\to\infty}\bigg[\frac{-2(b+2)}{e^{\frac{b}{2}}} + \frac{2(2)}{e^{\frac{1}{2}}} \bigg] \;\;=\;\;0 + \frac{4}{e^{\frac{1}{2}}} \;\;=\;\;\frac{4}{\sqrt{e}}


    The integral converges . . . Therefore, the series will converge.

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  3. #3
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    #1 - Integrate the function.

    You should arrive at -2x*e^(-x/2) -4e^(-x/2).

    Now evaluate the integral for the bounds lim N->infinity (1, N).
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