Use mathematical induction to prove: $$\frac{n}{2} < \sum_{k=1}^{2^n-1}\frac{1}{k} < n$$ To prove the left part. Assume: $$\frac{m}{2} < \sum_{k=1}^{2^m-1}\frac{1}{k}$$ Want to prove: $$\frac{m+1}{2} < \sum_{k=1}^{2^{(m+1)}-1}\frac{1}{k}$$ I have: $$\sum_{k=1}^{2^{(m+1)}-1}\frac{1}{k} = \sum_{k=1}^{2^{m}-1}\frac{1}{k} + \sum_{k=2^m}^{2*2^{m}-1}\frac{1}{k} > \frac{m}{2} + \sum_{k=2^m}^{2*2^{m}-1}\frac{1}{k}$$ It is enough to prove: $$\frac{m}{2} + \sum_{k=2^m}^{2*2^{m}-1}\frac{1}{k} > \frac{m+1}{2}$$ $$\frac{1}{2} < \sum_{k=2^m}^{2*2^{m}-1}\frac{1}{k} $$ From here I don't know how to continue or maybe I was wrong from the beginning. Can someone please tell me how to do it?
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2Each of the terms in the last sum is $>\frac1{2^{m1}}$ – saulspatz Mar 11 '21 at 21:35
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But in this case we get the last sum $ > \frac{1}{2^{m+1}} * (2^m - 1) = \frac{1}{2}-\frac{1}{2^{m+1}}$ and is smaller than $1/2$? – SixTwelfthsPi Mar 11 '21 at 22:19
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2How many numbers are there from $2^m$ to $2^{m+1}-1$ inclusive? – saulspatz Mar 11 '21 at 22:22
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Ohohoh sorry! $2^m$ it is.. I see the solution. Thank you!! – SixTwelfthsPi Mar 11 '21 at 22:25