How to proof that $\sum_{i=1}^{2^n} 1/i \ge 1+n/2$ I had troubles trying to prove that for every $n\ge1$
$$\sum_{i=1}^{2^n}\frac1i\ge 1+\frac n2$$
Can you give me a hint about the induction proof or show me in detail how can I prove it? I would appreaciate any help. Thanks!

$$\sum_{i=1}^{2^{k+1}}\frac{1}{i} = \sum_{i=1}^{2^{k}}\frac{1}{i}+\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i} \\ = \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^k+1}+\frac{1}{2^k+1} \ldots +\frac{1}{2^{k+1}-1}+ \frac{1}{2^{k+1}}\right) \\ \geq \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}} \ldots +\frac{1}{2^{k+1}}+ \frac{1}{2^{k+1}}\right)$$ The "greater than or equal to" is true because you are making the denominators larger. Next, you know you are adding up $2^k$ quantities of  $\frac{1}{2^{k+1}}$ since $2^{k+1}-2^{k} = 2^k(2-1) = 2^k$. Hence $$\sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}} \ldots +\frac{1}{2^{k+1}}+ \frac{1}{2^{k+1}}\right) = \sum_{i=1}^{2^{k}}\frac{1}{i}+2^k\left(\frac{1}{2^{k+1}}\right) \\ = \sum_{i=1}^{2^{k}}\frac{1}{i}+\frac{1}{2}$$ Now apply the induction hypothesis and do some algebra to get your result!
