Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.