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Math Help - monotonic decreasing

  1. #1
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    monotonic decreasing

    \sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}

    the dividing of a_n+1 /a_n is not conlusive
    ?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    \sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}} the dividing of a_n+1 /a_n is not conlusive?
    What exactly is the question? The title is almost meaningless.
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  3. #3
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    i need to prove that a_n is monotonickly decreasing
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i need to prove that a_n is monotonickly decreasing
    Can you show that \dfrac{x^2}{e^{x^2}} is decreasing for x\ge 1~?
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  5. #5
    Senior Member Sambit's Avatar
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    Arrow

    hint: the denominator increases more rapidly than the numerator.
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  6. #6
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    i can show a_n+1 -a_n <0

    or

    a_n+1 /a_n=q
    and show that q<1

    both i have written but i dont know how to conclude from the expression
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  7. #7
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    Quote Originally Posted by transgalactic View Post
    i can show a_n+1 -a_n <0

    or

    a_n+1 /a_n=q
    and show that q<1

    both i have written but i dont know how to conclude from the expression
    Now, you have me confused! a_{n+1}- a_n< 0 is the same as a_{n+1}< a_n and that is the definition of "decreasing".

    Or are you saying that you know you have to prove one of those but don't know how?

    I would look at \frac{a_{k+1}}{a_k}= \frac{(k+1)^2}{e^{(k+1)^3}}\frac{e^{k^3}}{k^2} = \frac{(k+1)^2}{k^2}e^{k^3-((k+1)^3}= \frac{(k+1)^2}{k^2}e^{-3k^2- 3k- 1}.

    Now, in the limit, as k goes to infinity, that fraction goes to 1 so this "eventually" decreases to 0.

    If you really need to prove that it is decreasing for all k greater than or equal to 1, I recommend Plato's suggestion. Take the dervative of x^2e^{-x^3}.
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