Inequality proof

  • #1
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Homework Statement


[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]?
 

Answers and Replies

  • #2
tiny-tim
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Hi jeff! :smile:

Hint: they're all greater than the last one. :wink:
 
  • #3
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Hi jeff! :smile:

Hint: they're all greater than the last one. :wink:

I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque
 
  • #4
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I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

If a < b < c
a + a + a < a + b + c

Right?

Try something along those lines.
 
  • #5
Dick
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I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

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?
 
  • #6
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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?

Yes I agree, but that isn't helping me at all- I just don't know how to formulate a proof for this.
 
  • #7
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There are [tex]2^{k+1}[/tex] terms
 
  • #10
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never minnd, I actually did this in a different problem- thanks everyone
 
  • #11
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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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