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Math Help - Infinite series of an exponent question.

  1. #1
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    Infinite series of an exponent question.

    I'm trying to evaluate the infinite series of e^Nx from N=0. As this expands to:
    1 + e^x + e^2x + ... e^Nx

    Can I treat it as a geometric series, where r = e^x, collapsing it to

    1/(1 - e^x)?
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  2. #2
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    Quote Originally Posted by DenJansen View Post
    I'm trying to evaluate the infinite series of e^Nx from N=0. As this expands to: 1 + e^x + e^2x + ... e^Nx
    If e^x\ge 1, As an infinite series that series diverges.
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  3. #3
    MHF Contributor Amer's Avatar
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    \displaystyle 1 + e^x + e^{2x} + ... e^{Nx}

    \displaystyle \left(1 + e^x + e^{2x} + ... e^{Nx}\right) \left(\frac{(e^x - 1 )}{e^x -1 }\right)

    \displaystyle \frac{e^{Nx+1} -1 }{e^x -1 }
    Last edited by Amer; March 10th 2011 at 12:34 PM.
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    Ah, I see. Thank you both!
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    Wait, I just noticed something. As this is an infinite series from N=0 to N->∞, wouldn't the equation above simply = ∞? Is there a way to solve this where N is not in the final equation?
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  6. #6
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    Quote Originally Posted by DenJansen View Post
    Wait, I just noticed something. As this is an infinite series from N=0 to N->∞, wouldn't the equation above simply = ∞? Is there a way to solve this where N is not in the final equation?
    I do not understand that question.
    You have two different ideas going on: e^x~\&~N.
    If x<0, that is 0<e^x<1, then the series does converge.
    Its sum is \dfrac{1}{1-e^x}.

    As I said in reply #2, if e^x\ge 1 the series diverges.

    So the answer depends upon the value of x.
    Whereas, N is simply an index for the sum.
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