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Mathematics 20 Online
OpenStudy (anonymous):

Determine whether the sequence is increasing, decreasing

OpenStudy (anonymous):

\[a_{n} = n e ^{-n}\]

OpenStudy (anonymous):

i can see tis is decreasing...but dunno how to prove it ....any idea ?

OpenStudy (anonymous):

Try To rewrite it..

OpenStudy (anonymous):

e^-n.... means 1\e^n

OpenStudy (nikvist):

\[\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)}}{ne^{-n}}=\frac{n+1}{e}\] \[a_2<a_1\]

OpenStudy (anonymous):

a2?

OpenStudy (anonymous):

where did you get a2 from lol? :)

OpenStudy (anonymous):

thanks...he mean a n+1

OpenStudy (anonymous):

oh, lol, atleast you got it ^_^

OpenStudy (anonymous):

thanks all.. ^ ^ how bout this question.. \[a _{n} = n ^{n} / n!\]

OpenStudy (anonymous):

ratio test ^_^

OpenStudy (anonymous):

i using the method as up there...but getting complicated as i implant it

OpenStudy (anonymous):

are you sure? it seems simple an+1 /an :) give it a one more try

OpenStudy (anonymous):

i get this.. \[a _{n+1} \div a _{n} = n+1 / n ^{n}\]

OpenStudy (anonymous):

no dear , you'll get something like this: an+ 1 = (n+1)^(n+1)/(n+1) ! then divide this with an, give it a try

OpenStudy (anonymous):

i getting that after simplified.. thsoe

OpenStudy (anonymous):

alright, let's do it the easier way, take points for n and see whether it's increasing or decreasing :)

OpenStudy (anonymous):

if the number is getting bigger then it's diverging, if the number is getting smaller then it's converging :)

OpenStudy (anonymous):

ok..my bad..i found the answer..thanks

OpenStudy (anonymous):

np ^_^

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