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anonymous

  • 5 years ago

Determine whether the sequence is increasing, decreasing

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  1. anonymous
    • 5 years ago
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    \[a_{n} = n e ^{-n}\]

  2. anonymous
    • 5 years ago
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    i can see tis is decreasing...but dunno how to prove it ....any idea ?

  3. anonymous
    • 5 years ago
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    Try To rewrite it..

  4. anonymous
    • 5 years ago
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    e^-n.... means 1\e^n

  5. nikvist
    • 5 years ago
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    \[\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)}}{ne^{-n}}=\frac{n+1}{e}\] \[a_2<a_1\]

  6. anonymous
    • 5 years ago
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    a2?

  7. anonymous
    • 5 years ago
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    where did you get a2 from lol? :)

  8. anonymous
    • 5 years ago
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    thanks...he mean a n+1

  9. anonymous
    • 5 years ago
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    oh, lol, atleast you got it ^_^

  10. anonymous
    • 5 years ago
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    thanks all.. ^ ^ how bout this question.. \[a _{n} = n ^{n} / n!\]

  11. anonymous
    • 5 years ago
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    ratio test ^_^

  12. anonymous
    • 5 years ago
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    i using the method as up there...but getting complicated as i implant it

  13. anonymous
    • 5 years ago
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    are you sure? it seems simple an+1 /an :) give it a one more try

  14. anonymous
    • 5 years ago
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    i get this.. \[a _{n+1} \div a _{n} = n+1 / n ^{n}\]

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    i getting that after simplified.. thsoe

  17. anonymous
    • 5 years ago
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    alright, let's do it the easier way, take points for n and see whether it's increasing or decreasing :)

  18. anonymous
    • 5 years ago
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    if the number is getting bigger then it's diverging, if the number is getting smaller then it's converging :)

  19. anonymous
    • 5 years ago
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    ok..my bad..i found the answer..thanks

  20. anonymous
    • 5 years ago
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    np ^_^

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