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. 

$$
\underbrace{ 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k} + \frac 1 {2^{k+1}} }_{\Large\text{This is wrong.}}
$$
$$
\overbrace{ \underbrace{ \underbrace{1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k}}_{\Large\text{The is the case }n=k.} + \frac 1 {2^k+1} + \frac 1 {2^k+2} + \frac 1 {2^k+3} + \cdots + \frac 1 {2^{k+1}} } }_{\Large\text{This is the case } n = k+1.}^{\Large\text{This is what you need instead.}}
$$
For example, suppose $n=k=3.$ Then
$$
1+ \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k} = 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7 + \frac 1 8.
$$
And the next case, $n=k+1 = 3+1=4$ is this
$$
\underbrace{ \underbrace{1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7 + \frac 1 8}_{\Large\text{This is the case } n=3.} + \frac 1 9 + \frac 1{10} + \frac 1 {11} + \frac 1 {12} + \frac 1 {13} + \frac 1 {14} + \frac 1 {15} + \frac 1 {16}.}_{\Large\text{This is the case } n = 4.}
$$
You don't just add one more term when you increment $n$ from $k$ to $k+1;$ you add $2^k$ terms to get up to $2^{k+1}.$
