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Math Help - infinit series

  1. #1
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    infinit series

    find the sum of the infinit series, if the it converges.

    infinity ((-1)^n-1(6n)) / (e^2n)
    E
    n = 1

    I'm confused because of the natural log in the denominator.
    does it converge or diverge?

    i just get errors while graphing.
    How can i prove this mathematically.

    thanks for the help.
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  2. #2
    Eater of Worlds
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    Is this it:

    \sum_{n=1}^{\infty}\frac{(-1)^{n}-6n}{e^{2n}}

    If so, it converges.


    Actually, it converges to:

    \frac{3}{2(sinh(1))^{2}}-\frac{1}{e^{2}+1}\approx{-1.20529541347}
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  3. #3
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    Almost had it, just forget the n-1 part, instead of just n

    here it is in image format:

    there ya go.
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  4. #4
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    Let's do it this way then:

    \sum_{n=1}^{\infty}\frac{(-1)^{n-1}6^{n}}{e^{2n}}

    Let S=\frac{6}{e^{2}}-\frac{6^{2}}{e^{4}}+\frac{6^{3}}{e^{6}}-\frac{6^{4}}{e^{8}}-..................

    Factor out \frac{6}{e^{2}}

    S=\frac{6}{e^{2}}\left(1-\underbrace{\frac{6}{e^{2}}+\frac{6^{2}}{e^{4}}-\frac{6^{3}}{e^{6}}-...............}_{\text{this equals S}}\right)

    Then we have:

    \frac{e^{2}}{6}S=1-S

    \left(\frac{e^{2}}{6}+1\right)S=1

    Solve for S and we find S=

    S=\frac{6}{e^{2}+6}

    Therefore, it converges to:

    \frac{6}{e^{2}+6}\approx{0.448127183549}
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  5. #5
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    Hello, rcmango!

    Galactus did his usual excellent job.
    . . We can make it a wee bit shorter . . .


    Find the sum of the infinite series: . \sum^{\infty}_{n=1}\frac{(\text{-}1)^{n-1}6^n}{e^{2n}}

    We have: . \frac{6}{e^2} - \frac{6^2}{e^4} + \frac{6^3}{e^6} - \frac{6^4}{e^8} + \hdots

    This is a geometric series with first term: a = \frac{6}{e^2} .and common ratio: r = \text{-}\frac{6}{e^2}

    Therefore, its sum is: . \frac{6}{e^2}\cdot\frac{1}{1 - \left(\text{-}\frac{6}{e^2}\right)} \;=\;\frac{6}{e^2+6}

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  6. #6
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    Ya, thanks very much galactus!

    You really broke it down nice so i could understand it fully.

    appreciated this very much.
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