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

the dividing of a_n+1 /a_n is not conlusive

?

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- Mar 5th 2011, 05:28 AMtransgalacticmonotonic decreasing
$\displaystyle \sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}$

the dividing of a_n+1 /a_n is not conlusive

? - Mar 5th 2011, 05:33 AMPlato
- Mar 5th 2011, 07:41 AMtransgalactic
i need to prove that a_n is monotonickly decreasing

- Mar 5th 2011, 07:45 AMPlato
- Mar 5th 2011, 09:01 AMSambit
hint: the denominator increases more rapidly than the numerator.

- Mar 5th 2011, 09:43 PMtransgalactic
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 - Mar 6th 2011, 03:42 AMHallsofIvy
Now, you have me confused! $\displaystyle a_{n+1}- a_n< 0$ is the same as $\displaystyle a_{n+1}< a_n$ 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 $\displaystyle \frac{a_{k+1}}{a_k}= \frac{(k+1)^2}{e^{(k+1)^3}}\frac{e^{k^3}}{k^2}$$\displaystyle = \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 $\displaystyle x^2e^{-x^3}$.