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Show if the sequence converges

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data
    showing it is increasing with an upper bound or decreasing
    with a lower bound

    n/(2^n)

    2. Relevant equations

    if an >an+1 its underbound
    an< an+1 then its upperbound

    3. The attempt at a solution
    I tried first by finding out the sequence:
    a1=1/2
    a2=1/2
    a3=3/8
    a4=1/4
    a5=5/32

    I'm assuming its decreasing, but I'm not sure if this is monotonic at all. Considering how a1=a2 and then a2>a3 and a3>a4 then a4>a5. I think it's underbound since an>an+1 but the first part threw me off since a1=a2. Someone clarify for me?
     
  2. jcsd
  3. May 15, 2010 #2

    LCKurtz

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    It is really trivial to show that sequence is decreasing. Have you tried?
     
  4. May 15, 2010 #3
    Yea I can tell that the sequence is decreasing just by looking at that. I was taught to try (an+1)-an see if that's greater than 0 or not, but its kinda confusing with this equation. then there is also taking the derivative but it also doesn't help much because that ends up being just 1/2^n-1/2^n(log2)
     
  5. May 15, 2010 #4

    LCKurtz

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    That's good advice. What happens if you work on the inequality an+1 < an with reversible steps? Try putting the 2's on one side by themselves.
     
  6. May 15, 2010 #5
    Ummm... do you think you could give me an example as to how to go about this. I have other problems and this one seems the most simple to actually work with. >.<;; my professor doesn't give good examples and that's usually how I work through my math. Lets see if I can get this right

    so n+1/(2n+1< n/2n
    That turns into n+1/n < 2n+1/2n If i subtract that...
    so...
    n+1/n-2n+1/2n<0 so that shows that its decreasing and underbounded (meaning its bounded under something right?

    but wait if i do it the other way... isn't 0>2n+1/2n-n+1/n
     
    Last edited: May 15, 2010
  7. May 16, 2010 #6

    LCKurtz

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    You need parentheses around the n+1 on the left don't you? Don't subtract anything. Just look at what you have. Simplify the right side. Compare it to the left.
     
  8. May 17, 2010 #7
    how do i simplify 2n+1/1n? does it become 1 or something?
     
  9. May 17, 2010 #8

    LCKurtz

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    [tex]\frac{2^{n+1}}{2^n}\neq\frac{2^{n+1}}{1^n}[/tex]
     
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