Question on the proof of divergence of harmonic series. The harmonic series, $\sum_{k=1}^{\infty}\frac{1}{k}$, diverge. Let the partial sum be denoted as $s_n = \sum_{k=1}^{n}\frac{1}{k}$. If $s_n$ was convergent, then it would have to be bounded and its subsequence $s_{2^n}$ must be also bounded. But, $$s_{2^n} = (1) + (\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+\dots+(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n})>1+\frac{n}{2}$$ as $n \rightarrow \infty$. I can see that each term in parenthesis are greater than $\frac{1}{2}$. But the book doesn't show how $(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}) > \frac{1}{2}$ for each $n$. Is this supposed to be intuitively obvious or is there some way to prove this rigorously?

$$\left(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}\right) > \underbrace{\left(\frac{1}{2^n}+\dots+\frac{1}{2^n}\right)}_{2^{n-1} \text{times}} = \frac{2^{n-1}}{2^n} = \frac{1}{2} $$
