sequence monotonic? bounded?

**twilightstr** · March 14th 2009, 11:20 PM

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?

**o_O** · March 14th 2009, 11:25 PM

Consider $f(x) = xe^{-x} > 0$ for all $x > 0$

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

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

So we have our two bounds.

**twilightstr** · March 14th 2009, 11:30 PM

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

**HallsofIvy** · March 15th 2009, 04:16 AM

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

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

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