Proof by induction, dont know how to represent range The question asks for me to prove the following through induction:
$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} \geq 1 + \frac{n}{2}$
This is my proof thus far: 
Proving true for $n = 1$
\begin{align*}
1 + \frac{1}{2} &\geq 1 + \frac{1}{2}\\
\end{align*}
Assuming true for $n = k$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} \geq 1 + \frac{k}{2}
\end{align*}
Proving true for $n = k + 1$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} &\geq 1 + \frac{k}{2} + \frac{1}{2}\\
(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2^{k+1}} &\geq (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2}\\
\frac{1}{2^{k+1}} &\geq \frac{1}{2}\\
\end{align*}
I realized at the last statement that something was off, because the last statement contradicts the thing I'm trying to prove. I then realized that it was because the difference between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$ sets was not simply the addition of $\frac{1}{2^{k + 1}}$, but rather, all numbers between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$.
For example n = 1 is $1 + \frac{1}{2}$, but $n = 2$ is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. (note the addition of $\frac{1}{3}$.)
So how can I represent this range?
I believe my proof works except for the fact that $\frac{1}{2^{k + 1}}$ needs to be replaced with something else, I just don't know what that is. 

Start with $$1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} \ge 1+\frac{k}{2}.$$
On the left-hand side, you want to add $\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}$ (all the terms from $2^k+1$ to $2^{k+1}$), as you noted. On the right-hand side you want to add $\frac{1}{2}$, as you have done.
It remains to show
$$\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} \ge \frac{1}{2}.$$
Can you do this?
Hint: each term on the left-hand side is larger than $\frac{1}{2^{k+1}}$, and there are $2^k$ terms.
