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Inequality proof

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{1}{2^{k}+1} + \frac{1}{2^{k} +2} + ... + \frac{1}{2^k + 2^k} \geq \frac{1}{2}[/tex]



    2. The attempt at a solution
    Not too sure, I am working on a larger proof (not too much difficult) and the above is my attempt to prove the induction step k+1 (since [tex]\frac{1}{2^k + 2^k} = \frac{1}{2^{k+1}}[/tex]).

    Should i try to factor out [tex]\frac{1}{2^k}[/tex]?
     
  2. jcsd
  3. Feb 21, 2010 #2

    tiny-tim

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    Hi jeff! :smile:

    Hint: they're all greater than the last one. :wink:
     
  4. Feb 21, 2010 #3
    I still don't see it, i wish it were clear to me.

    Thanks though,

    Jeffrey Levesque
     
  5. Feb 21, 2010 #4
    If a < b < c
    a + a + a < a + b + c

    Right?

    Try something along those lines.
     
  6. Feb 21, 2010 #5

    Dick

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    Don't you agree with tiny-tim that the last term in the sum is smaller than the rest? How many terms in the sequence are there?
     
  7. Feb 21, 2010 #6
    Yes I agree, but that isn't helping me at all- I just don't know how to formulate a proof for this.
     
  8. Feb 21, 2010 #7
    There are [tex]2^{k+1}[/tex] terms
     
  9. Feb 21, 2010 #8
    Are you sure? Count again. : )
     
  10. Feb 21, 2010 #9
    [tex]2^k[/tex] terms.
     
  11. Feb 21, 2010 #10
    never minnd, I actually did this in a different problem- thanks everyone
     
  12. Feb 21, 2010 #11
    I also noticed that the last term on the LHS is the smallest, but I was confused for awhile since I thought the denominators were 2^k + 2^0, 2^k + 2^1, ..., 2^k + 2^k, in which case there are k+1 terms, and the estimate fails.

    Anyways assuming the progression is what everyone else thinks it is, then basically [itex]2^i \leq 2^k[/itex] for i = 1, 2, ..., k so
    [tex]2^k + 2^i \leq 2^k + 2^k = 2^{k+1} \Rightarrow \frac{1}{2^k + 2^i} \geq \frac{1}{2^{k+1}} [/tex]
    for i = 1, 2, ..., k.
     
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