Summation Proof prove that: 
I am having trouble coming up with a rigorous proof for this.Can I use induction? Can anyone demonstrate? 
$$\sum_{i={2^n+1}}^{{2^{n+1}}}{1/i}\ge 1/2$$

$$
\sum_{i = 2^n + 1}^{2^{n+1}} \frac{1}{i} = \frac{1}{2^n + 1} + \frac{1}{2^n + } + \cdots + \frac{1}{2^n + 2^n}
$$
$$
> \frac{1}{2^n + 2^n} + \frac{1}{2^n + 2^n} + \cdots + \frac{1}{2^n + 2^n} = \frac{2^n}{2^n + 2^n} = \frac{1}{2}
$$
