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How to show that for $n\geqslant 2$ $$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}<n$$

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It's not true for $n=0,1$, you might want to specify for what $n$ – user50407 Dec 29 '12 at 17:52
fair enough. I just wish comments no longer applying would be deleted. – amWhy Dec 29 '12 at 17:59
So, the sum in the center is to be interpreted $$ \sum_{1 \le k \le 2^{n}-1}\frac{1}{k}, $$ where $n\ge 2$? – 000 Dec 29 '12 at 18:05
up vote 1 down vote accepted

You can group the sum as follows: $$\sum_{k=1}^{2^n}{\frac{1}{k}}> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}\right)...$$ So that: $$\sum_{k=1}^{2^n-1}{\frac{1}{k}} > 1 + \frac{n}{2}-\frac{1}{2^n}>\frac{n}{2}$$ If instead you group them by the lower power of two you easily get the other limit.

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HINT: $$\frac12<2^n\left(\frac1{2^{n+1}-1}\right)<\sum_{k=2^n}^{2^{n+1}-1}\frac1k<2^n\left(\frac1{2^n}\right)=1$$

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