Prove that $\sum_{i=2^n+1}^{2^{n+1}} \frac{1}{i} \ge 0.5$

Question

Where $n$ is any positive integer.

I'm honestly completely at a loss at how to prove this.
Tested by brute forcing it up to large numbers, and it keeps increasing, although very slowly.

This is actually part of a bigger problem containing harmonic numbers, but I've solved the rest, so that's why I don't really have much of an idea on how to approach this.
I've tried integrals to find something that would be smaller than the sum, yet clearly bigger than $0.5$, but, again, I can't really prove it.

Answer 1

Since $i\leq 2^{n+1}$ it follows that $\frac{1}{i}\geq \frac{1}{2^{n+1}}$. There are $2^n$ terms in the sum, hence $$ \sum_{i=2^n+1}^{2^{n+1}}\frac{1}{i}\geq \frac{2^n}{2^{n+1}}=\frac{1}{2}$$

Answer 2

$$\sum_{i=2^n+1}^{2^{n+1}} \frac1i \geq \sum_{i=2^n+1}^{2^{n+1}} \frac{1}{2^{n+1}} = \frac12.$$