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Math Help - Integral Test

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    Integral Test

    Does this fail the Integral Test? I think it does.

    Summation n=1 to infinity (ne^(-n/2))
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    Quote Originally Posted by saiyanmx89 View Post
    Does this fail the Integral Test? I think it does.

    Summation n=1 to infinity (ne^(-n/2))
    \int_{1}^{\infty}xe^{-\frac{x}{2}}dx

    using integration by parts ( and L'hosiptals rule)you get

    and please forgive my abuse of notation

    \int_{1}^{\infty}xe^{-\frac{x}{2}}dx=-\frac{x}{2}e^{-\frac{x}{2}}\bigg|_{1}^{\infty}-\int_{1}^{\infty}-\frac{1}{2}e^{-\frac{x}{2}}dx=\frac{e^{-\frac{1}{2}}}{2}+\frac{1}{2}\int_{1}^{\infty}e^{-\frac{x}{2}}dx

    Since the latter integral also coverges the series coverges by the integral test.
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  3. #3
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    Helo, saiyanmx89!

    Does this fail the Integral Test? I think it does. . \sum^{\infty}_{n=1} n\,e^{\text{-}\frac{n}{2}}
    I believe it passes . . .


    We have: . \int^{\infty}_1 x\,e^{\text{-}\frac{x}{2}}dx

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

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

    Then: . \lim_{b\to\infty}\bigg[\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{6}{e^{\frac{1}{2}}}\bigg] \;=\;0 + \frac{6}{e^{\frac{1}{2}}}


    Therefore, the integral converges to: . \frac{6}{\sqrt{e}}

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