# Thread: sequence monotonic? bounded?

1. ## 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?

2. 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.

3. what do u mean by "we have our two bounds"
dont really get what bounded means. how is 1/e bounded?

4. Originally Posted by twilightstr
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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### Determine whether sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? an = ne

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