Prove by induction that $\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$

Question

As the title says I need to prove the following by induction: $$\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$$

When trying to prove that P(n+1) is true if P(n) is, then I get that:

$$\sum_{i=1}^{2^{n+1}} \frac{1}{i} = (\sum_{i=1}^{2^n} \frac{1}{i} ) +(\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} )$$ using the inductive hypothesis. And for P(n+1) to be true it would be sufficient to prove that $$1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n+1}{2}$$ or rewriting $$\sum_{i=1}^{2^n} \frac{1}{2^n+i} \ge \frac{1}{2} $$ and here I'm a bit stuck

Thank you for your help!

Answer 1

Providing this to make the inductive argument more explicit (some of the details were discussed in the comments): \begin{align} \sum_{i=1}^{2^{k+1}}\frac{1}{i}&= \sum_{i=1}^{2^k}\frac{1}{i}+\sum_{i=1}^{2^k}\frac{1}{2^k+i}\tag{rewrite the sum}\\[1em] &\geq 1+\frac{k}{2}+\sum_{i=1}^{2^k}\frac{1}{2^k+i}\tag{by inductive hypothesis}\\[1em] &\geq 1+\frac{k}{2}+\sum_{i=1}^{2^k}\frac{1}{2^{k+1}}\tag{since $i\leq2^n$}\\[1em] &= 1+\frac{k}{2}+2^k\left(\frac{1}{2^{k+1}}\right)\tag{simplify}\\[1em] &= 1+\frac{k}{2}+\frac{1}{2}\tag{simplify}\\[1em] &= 1+\frac{k+1}{2}\tag{simplify}. \end{align}

Answer 2

See Inequality of finite harmonic series

Also proving $\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$ clear shows that $1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n+1}{2}$ is true since adding $1$ to the LHS and $\frac{1}{2}$ to the RHS holds the inequality.