Determine whether the sequence is increasing, decreasing

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Determine whether the sequence is increasing, decreasing

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[a_{n} = n e ^{-n}\]
i can see tis is decreasing...but dunno how to prove it ....any idea ?
Try To rewrite it..

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Other answers:

e^-n.... means 1\e^n
\[\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)}}{ne^{-n}}=\frac{n+1}{e}\] \[a_2
a2?
where did you get a2 from lol? :)
thanks...he mean a n+1
oh, lol, atleast you got it ^_^
thanks all.. ^ ^ how bout this question.. \[a _{n} = n ^{n} / n!\]
ratio test ^_^
i using the method as up there...but getting complicated as i implant it
are you sure? it seems simple an+1 /an :) give it a one more try
i get this.. \[a _{n+1} \div a _{n} = n+1 / n ^{n}\]
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
i getting that after simplified.. thsoe
alright, let's do it the easier way, take points for n and see whether it's increasing or decreasing :)
if the number is getting bigger then it's diverging, if the number is getting smaller then it's converging :)
ok..my bad..i found the answer..thanks
np ^_^

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