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Math Help - geometric series

  1. #1
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    geometric series

    given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.

    a) set x=1 in the above series and write out the first 4 terms
    b) find the exact sum of the series s(1)
    c) fing the exact sum of the series s(x)
    d) for what values of x does the series from part (c) converge?

    for part A i set the S(1) = the sum from n=1 to infinity of (-1/e)^n, and i think i got the first 4 terms but i still don't know what it converges to.

    so on part B i got a sum of s(1) = e/(e+1) is it right?

    on part C i got S(x) = e^x/(e^x+1) right?

    since i am not sure about my answers i don't know part D, for what values it converges.

    help me please
    thanks
    K
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by kithy View Post
    given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.

    a) set x=1 in the above series and write out the first 4 terms
    b) find the exact sum of the series s(1)
    c) fing the exact sum of the series s(x)
    d) for what values of x does the series from part (c) converge?

    for part A i set the S(1) = the sum from n=1 to infinity of (-1/e)^n, and i think i got the first 4 terms but i still don't know what it converges to.
    You're not asked that... Your series is right.

    so on part B i got a sum of s(1) = e/(e+1) is it right?
    Right

    on part C i got S(x) = e^x/(e^x+1) right?
    Only if e^{-x}<1 (else, the series diverges), so it's for x>0 (this is part D).

    Actually, the exact value of s(x) is : \sum_{k=1}^{\infty} (-e^{-x})^n=\lim_{N \to \infty} \frac{1+(\frac{1}{e^x})^N}{1+\frac{1}{e^x}} (simplify it the way you want)

    since i am not sure about my answers i don't know part D, for what values it converges.
    You can use the alternate series test
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi

    Quote Originally Posted by kithy View Post
    given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.
    so on part B i got a sum of s(1) = e/(e+1) is it right?

    on part C i got S(x) = e^x/(e^x+1) right?
    I don't agree.

    s(1)=\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^n is negative because it's an alternating series and \left(-\frac{1}{\mathrm{e}}\right)^1=-\frac{1}{\mathrm{e}}<0 so s(1)=\frac{\mathrm{e}}{1+\mathrm{e}} can't be true.

    You may try rewriting the series as :

    s(1)=\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^n=-\frac{1}{\mathrm{e}}\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^{n-1}<br />
=-\frac{1}{\mathrm{e}}\underbrace{\left(1-\frac{1}{\mathrm{e}}+\left(-\frac{1}{\mathrm{e}}\right)^2+\left(-\frac{1}{\mathrm{e}}\right)^3+\ldots\right)}_{\fra  c{1}{1-\left(-\frac{1}{\mathrm{e}}\right)}}<br />
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  4. #4
    Moo
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    Ow, this is the mistake :

    The sum of terms of a geometric series is that thing, multiplied by the first term.
    Here, the first term is -\frac 1e

    Hence, s(1)=-\frac 1e \cdot \frac{e}{1+e}=-\frac{1}{1+e}

    sorry about that
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