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Thread: sequence monotonic? bounded?

  1. #1
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    sequence monotonic? bounded?

    is an= ne^-n increasing or decreasing? i think the sequence decreases for n greater than 1 and increases for n less than 1 but i only pay attention the the sequence as n goes to infitny right? also is the sequence bounded? if so how can i tell? and generally, how do i know if the sequence is bounded above?
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  2. #2
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    Consider $\displaystyle f(x) = xe^{-x} > 0$ for all $\displaystyle x > 0$

    We have: $\displaystyle f'(x) = e^{-x} - xe^{-x} < 0$ for $\displaystyle x > 1$

    So we know $\displaystyle f(x)$ (and thus $\displaystyle f(n) = a_n$) is decreasing after $\displaystyle n > 1$ where $\displaystyle f(1) = \frac{1}{e}$

    So we have our two bounds.
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  3. #3
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    what do u mean by "we have our two bounds"
    dont really get what bounded means. how is 1/e bounded?
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  4. #4
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    Quote Originally Posted by twilightstr View Post
    what do u mean by "we have our two bounds"
    dont really get what bounded means. how is 1/e bounded?
    Upper and lower bounds.

    "n" is a positive integer and $\displaystyle e^x$ is always positive so $\displaystyle ne^{-n}> 0$ which says that 0 is a "lower bound".

    $\displaystyle f(x)= x e^{-x}$ is a decreasing function, as o_O showed, so for all n, $\displaystyle ne^{-n}< e^{-1}$ and so $\displaystyle e^{-1}$ is an "upper bound" on the sequence.
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