I have to use induction to prove the following inequality:
$\sum_{k=1}^{2^{n}} \frac{1}{k} \geq 1 + \frac{n}{2}$.
So the base case is $n=1$ and it's true because both sides will equal $\frac{3}{2}$.
If we assume $n=m+1$ we get
$\sum_{k=1}^{2^{m+1}} \frac{1}{k} \geq 1 + \frac{m+1}{2}$.
But I'm not sure how to prove that the LHS=RHS.
Is it true to say this:
$\sum_{k=1}^{2^{m}} \frac{1}{k} + \frac{1}{m+1} \geq 1 + \frac{m+1}{2}$?
To finish the proof it is enough to note that
$$\sum\limits_{k=1}^{2^{m+1}} \dfrac{1}{k} = \sum\limits_{k=1}^{2^{m}} \dfrac{1}{k} + \sum\limits_{k=2^{m} + 1}^{2^{m+1}} \dfrac{1}{k}$$
The first part of the sum in the right side is $\geqslant 1+\dfrac{m}{2}$, the other one is $\geqslant \sum\limits_{k=2^{m} + 1}^{2^{m+1}} \dfrac{1}{k} \geqslant 2^m \cdot \dfrac{1}{2^{m + 1}} = \dfrac{1}{2}$, so, overall, $$\sum\limits_{k=1}^{2^{m+1}} \dfrac{1}{k} \geqslant 1+ \dfrac{m + 1}{2}$$ which completes your proof by induction.