<s>[INST] Use induction to prove that $\sum_{k=1}^{2^{n}}\frac{1}{k} \geq 1 +\frac{n}{2}$? For the base case I have $n = 1$, so 
${1\over1} + {1\over2} = {3\over2}, 1 + {1\over2} = {3\over2}$. Hence this is true. For the induction hypothesis we can assume n=k is true also. For the induction step I plug in $n=k+1$ but i'm not sure what to do with it. 
 [/INST] If $n=1$, then
$$
\sum_{k=1}^{2^{n}}\frac{1}{k} = 1 + \frac{1}{2} = 1 + \frac{n}{2};
$$
if $n \geq 1$ is an integer such that
$$
\sum_{k=1}^{2^{n}}\frac{1}{k} \geq 1 + \frac{n}{2},
$$
then
\begin{align*}
\sum_{k=1}^{2^{n+1}}\frac{1}{k} 
&= \sum_{k=1}^{2^{n}}\frac{1}{k} + \sum_{k=2^{n}+1}^{2^{n+1}}\frac{1}{k}\\
&\geq 1 + \frac{n}{2} + \sum_{k=2^{n}+1}^{2^{n+1}}\frac{1}{k}\\
&= 1 + \frac{n}{2} + \frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}}\\
&\geq
1 + \frac{n}{2} + \frac{(2^{n+1} - 2^{n})}{2^{n}+1}\\
&= 1 + \frac{n}{2} + \frac{2^{n}}{2^{n}+1}\\
&\geq 1 + \frac{n}{2} + \frac{2^{n}}{2\cdot 2^{n}}\\
&= 1 + \frac{n}{2} + \frac{1}{2}\\
&= 1 + \frac{n+1}{2}.
\end{align*}
</s>