<s>[INST] Proving inequality $\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + + \frac{1}{2^{n}-1} < n$ Prove the inequality $\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + + \frac{1}{2^{n}-1} < n$ where $n\in\mathbb{N}\backslash\{0,1\}$
My work
At first I tried to prove first half by induction

$\frac{n}2+\frac{1}2<1+\frac{1}2+\dots+\frac{1}{2^{n+1}-1}$
Which is $\displaystyle\frac{1}2+\sum_{k=0}^{2^n-1}\frac{1}k<\sum_{k=0}^{2^{n+1}-1}\frac{1}k$
And then I got this $\displaystyle\sum_{k=0}^{2^{n+1}-1}\frac{1}k-\sum_{k=0}^{2^n-1}\frac{1}k=\sum_{k=0}^{2n}\frac{1}{2^n+k}$
And now I see no way to go on $\displaystyle\sum_{k=0}^{2n}\frac{1}{2^n+k}>\frac{1}2$
where $(n>1,n\in\mathbb{N},h=\overline{0,2n})$

My math background is too low so I'm not able to prove this last inequality I've got, maybe someone knows how to?(maybe using limit's definition?) But I think there should be more simplicated solution, which I cannot find for almost 4 hours now. Thank you
 [/INST] To prove the first inequality, group them $$1+(\frac12)+(\frac13+\frac14)+(\frac15+\frac16+\frac17+\frac18)+...$$
so each group goes from one power of two to the next.  Show that each group's sum is more than $1/2$.
To prove the second, group them $$1 + (\frac12+\frac13)+(\frac14+\frac15+\frac16+\frac17)+...$$
again, each group goes from one power of two to the next.  Each group's sum is less than $1$.
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