monotonic decreasing

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March 5th 2011, 05:28 AM
transgalactic

monotonic decreasing

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

the dividing of a_n+1 /a_n is not conlusive
?
March 5th 2011, 05:33 AM
Plato

Quote:

Originally Posted by transgalactic

$\sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}$ the dividing of a_n+1 /a_n is not conlusive?

What exactly is the question? The title is almost meaningless.
March 5th 2011, 07:41 AM
transgalactic

i need to prove that a_n is monotonickly decreasing
March 5th 2011, 07:45 AM
Plato

Quote:

Originally Posted by transgalactic

i need to prove that a_n is monotonickly decreasing

Can you show that $\dfrac{x^2}{e^{x^2}}$ is decreasing for $x\ge 1~?$
March 5th 2011, 09:01 AM
Sambit

hint: the denominator increases more rapidly than the numerator.
March 5th 2011, 09:43 PM
transgalactic

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
March 6th 2011, 03:42 AM
HallsofIvy

Quote:

Originally Posted by transgalactic

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 $x^2e^{-x^3}$ .