Proving this inequality by mathematical induction

Question

I want to prove this inequality:

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

So, if I suppose that the inequality holds for a natural number $k$, then

$$ \sum\limits_{i=1}^{2^{k+1}} \frac{1}{i} = {\sum\limits_{i=1}^{2^k} \frac{1}{i}} + {\sum\limits_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}}.$$

Thus, I just have to prove that $\sum\limits_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i} \geq \frac{1}{2}$, but I'm stuck on this. I know there are $2^{k}$ natural numbers bewteen $2^k$ and $2^{k+1}$, but I'm not sure how to use it . I'd appreciate your help.

Answer 1

It's enough to prove that $$1+\frac{n}{2}+\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^{n+1}}\geq1+\frac{n+1}{2}$$ or $$\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^{n+1}}\geq\frac{1}{2},$$ which is true because $$\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^{n+1}}\geq\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+...+\frac{1}{2^{n+1}}=\frac{2^n}{2^{n+1}}=\frac{1}{2}.$$

Answer 2

Hint Observe that $1/i\ge 1/2^{k+1}$ for $i=2^{k}+1,\cdots, 2^{k+1}$.

Answer 3

\begin{align} \sum_{i=1}^{2^{k+1}}&=\sum_{i=1}^{2^{k}}\frac{1}{i}+\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i}\\ &\geq 1+\frac{k}{2} +\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i}\\ &\geq 1+\frac{k}{2} +\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{2^{k+1}}\\ &=1+\frac{k}{2} +\frac{2^{k+1}-2^k}{2^{k+1}}\\ &= 1+\frac{k+1}{2} \end{align}