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Math Help - finding the limit of the sequence

  1. #1
    Senior Member
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    finding the limit of the sequence

    finding this limit:

    lim
    n -> infinity (e^-n(-3)^n)

    After trying to graph this, i just get errors.
    I assume its because i tried to raise a natural log to a negative number?

    i imagine that this does not have a limit?
    how can i prove this mathematically.

    thankyou.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by rcmango View Post
    finding this limit:

    lim
    n -> infinity (e^-n(-3)^n)

    After trying to graph this, i just get errors.
    I assume its because i tried to raise a natural log to a negative number?

    i imagine that this does not have a limit?
    how can i prove this mathematically.

    thankyou.
    I assume you mean:

     \lim_{n \to \infty} e^{-n}\,(-3)^n

    Rearrange this:

    \lim_{n \to \infty} e^{-n}\,(-3)^n=\lim_{n \to \infty} (-1)^n e^{-n}\,(3)^n = \lim_{n \to \infty}(-1)^n e^{-n}\,e^{n\,\ln(3)} =\lim_{n \to \infty}(-1)^n e^{-n+n\,\ln(3)}

    ..... =\lim_{n \to \infty}(-1)^n e^{(\ln(3)-1)n}

    Now the coefficient of n in the exponential is >0 so the exponential term goes to infinity, while the powers of -1 change the sign of alternate terms. So the absolute value of this sequence goes to infinity, but with indeterminate sign, that is there is no limit.

    RonL
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  3. #3
    Senior Member
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    very nice, Thankyou.
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