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

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*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 $n= 1$


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

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