Prove by induction if $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ then $H_{2^n}\ge 1+\frac{n}{2}$?

Question

Suppose $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ and we want to prove that $H_{2^n}\ge 1+\frac{n}{2}$?

1. Since $\sum\limits_{i=1}^{2^1}\frac{1}{i}\ge1+\frac{1}{2}$ note that $H_{2^n}\ge 1+\frac{n}{2}$ for $n= 1$

2. Assume $H_{2^n}\ge1+\frac{n}{2}$ for all $n\le k$.

Note $$\sum\limits_{i=1}^{2^{n}}\frac{1}{i}\ge1+\frac{n}{2}$$

However, I'm having trouble showing $\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \frac{1}{2}$ inorder to show that

$$H_{2^{n+1}}=\left(\sum\limits_{i=1}^{2^n}\frac{1}{i}\right)+\left(\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\right)\ge \left(1+\frac{n}{2}\right)+\left(\frac{1}{2}\right)=1+\frac{(n+1)}{2}$$

How do we show by induction that:

$\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \frac{1}{2}$

Is there a better method to prove by induction that:

$\sum\limits_{i=1}^{2^{n+1}}\frac{1}{i}\ge \frac{(n+1)}{2}+1$

Answer 1

You see $$\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{2^{n+1} } = \frac{1}{2}$$