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nate h
nate h asked in Science & MathematicsMathematics · 9 years ago

How to find the sum of e^(-2k) (from k=0 to inf)?

I realize this series is converging and geometric.. but how do I get it into the form a/ (1-r)?

I'm just a bit confused on how to find the sums of some of these more interesting series.

4 Answers

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  • dennis
    Lv 6
    9 years ago
    Favorite Answer

    GP with first term e^0 = 1, common ratio e^-2

    so Sum to infinity is 1/ (1 - e^-2) = e^2/(e^2 -1).

  • Anonymous
    9 years ago

    If k increases, i.e. from 1 to 2, e^(-2k) decreases from e^-2 to e^-4, so 'r' is 1/e^2, with a=1.

    Therefore, Sn=a/(1-r)=1/[1-(e^-2)] which is some number like 1.1565176...

  • Jeff Aaron
    Lv 7
    9 years ago

    a = t(1) = e^(-2*0) = e^0 = 1

    t(2) = e^(-2*1) = e^(-2)

    r = t(2) / t(1) = e^(-2) / 1 = e^(-2)

    S = a / (1 - r)

    S = 1 / (1 - e^(-2))

    S =~ 1.1565176427496656518180806234654

  • Anonymous
    9 years ago

    .

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