Proving Inequality: 1/2^k+1 + 1/2^k+2 + ... + 1/2^(k+1) > 1/2

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SUMMARY

The inequality \(\frac{1}{2^{k+1}} + \frac{1}{2^{k+2}} + ... + \frac{1}{2^{k+1}} > \frac{1}{2}\) is proven by recognizing that each term \(\frac{1}{2^{k+1}}, \ldots, \frac{1}{2^{k+1}}\) is greater than \(\frac{1}{2^{k+1}}\). With \(2^k\) terms in total, the sum exceeds \(\frac{2^k}{2^{k+1}} = \frac{1}{2}\). This establishes the inequality definitively.

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Homework Statement



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[tex]\frac{1}{2^k+1}+\frac{1}{2^k+2}+...+\frac{1}{2^{k+1}}>\frac{1}{2}[/tex]

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The Attempt at a Solution



I cannot figure this out. It is part of a larger proof that I am trying to understand. Any help would be appreciated!
 
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Each of the terms [tex]\frac{1}{2^k+1},... \frac{1}{2^{k+1}-1}[/tex] is larger than [tex]\frac{1}{2^{k+1}}[/tex]

There are 2*2^k - 2^k = 2^k terms so the whole sum is certainly larger than 1/2:

[tex]\frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}} > \frac{2^k}{2^{k+1}} = \frac{1}{2}[/tex]
 

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