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Prove monotonicity

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data

    prove that the sequence (1+1/n)^(n+1) is monotonic decreasing

    2. Relevant equations



    3. The attempt at a solution
    I tried to use the binom expantion and the identity( Ck-1,n)+(Ck,n)=(Ck,n+1)
     
  2. jcsd
  3. Dec 4, 2012 #2
    All you have to do is show that a[itex]_{n}[/itex] ≥ a[itex]_{n+1}[/itex]
     
  4. Dec 4, 2012 #3
    obviously,how do i do that?
     
  5. Dec 4, 2012 #4
    When in doubt, use induction.
     
  6. Dec 4, 2012 #5
    If someone has an idea i will be glad to be helped.
     
  7. Dec 4, 2012 #6
    I'm serious. Use induction. Prove first that (1+1/1)^(1+1) is greater than or equal to (1+1/2)^(2+1) as your base case.

    I'm thinking about this problem and I think there actually is a better way of doing it but induction would be my first approach.
     
    Last edited: Dec 4, 2012
  8. Dec 4, 2012 #7

    Dick

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    If you want another approach look at the function f(x)=(1+1/x)^(x+1). Form the log and then take the derivative and try to show it's negative for x>=1.
     
  9. Dec 5, 2012 #8
    I search for a solution concerning sequences only without using functions.
     
  10. Dec 5, 2012 #9

    micromass

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    Most results using sequences can be proven most quickly using functions. Is there a special reason that you don't want to use functions??

    Anyway, you need to prove

    [tex]\left(1+\frac{1}{n}\right)^{n+1}\geq \left(1+\frac{1}{n+1}\right)^{n+2}[/tex]

    Can you algebraically manipulate the above to something suitable?
     
  11. Dec 5, 2012 #10
    I tried the binom expansion but it was not efficient.
     
  12. Dec 5, 2012 #11

    micromass

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    Try something else then. Try to make [itex]1+\frac{1}{n}[/itex] and the other one into one fraction. Then rearrange stuff.
     
  13. Dec 6, 2012 #12
    Up to now nothing works here.
     
  14. Dec 6, 2012 #13

    micromass

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    There are already 12 posts in this thread. I have seen many good suggestions here. But I have yet to see an attempt of you. If you say that "nothing works", then what did you try?? Can you show us exactly what you tried?? Why doesn't it work.

    If you expect to be spoonfed, then this thread will be locked.
     
  15. Dec 6, 2012 #14
    Try to solve and see the problem for yourself.This is not so easy as the increasing sequence in answer #9.
     
  16. Dec 6, 2012 #15

    micromass

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    I did solve it myself.
    So please make an attempt or this will be locked. If you say that nothing works, then surely there must be something that you tried?
     
  17. Dec 6, 2012 #16
    I need a solution without using functions and derivatives.I tried the binom expansion,the Bernouli inequality and more.the binom expansion may by efficient but i suspect that there must be a simpler way.
     
  18. Dec 6, 2012 #17

    micromass

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    Why did you ignore my post 11?
    There is a very simple way. You need to prove the inequality in post 9. First put
    [tex]1+\frac{1}{n}~~\text{and}~~1+\frac{1}{n+1}[/tex]
    into one fraction. Then rearrange the inequality in post 9. Play around with it. What do you get?
     
  19. Dec 6, 2012 #18
    solved.
     
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