Simplifying a summation - stuck after change of summation index I'm tracing through solution for a question I was working on. I don't quite understand how they got to the line which I marked with an arrow (apologies for using an image, I don't have much time left and didn't wanna have to look up how to do sigma notation in LaTeX here)

Text from the image:

The given sum is 
  $$\begin{align*}
&7\sum_{k=2}^n \frac1{k-1}-3\sum_{k=2}^n\frac1k-4\sum_{k=2}^n\frac1{k+1}\\
=&7\sum_{k=1}^{n-1} \frac1k-3\sum_{k=2}^n\frac1k-4\sum_{k=3}^{n+1}\frac1k\\
\color{red}{\longrightarrow}\overset{?}=&7\left(1+\frac12\right)-3\left(\frac12+\frac1n\right)-4\left(\frac1n+\frac1{n+1}\right)\\
&=9-\frac7n-\frac4{n+1}
\end{align*}$$

So I understand how they got the second line, which is just basically taking the coefficients out and using the change of summation index rule. 
What I don't understand is the line marked with the red arrow. 
I can see that $1$ comes from $1/1$ when $k=1$ and $1/2$ probably comes from $k=2$ but the summation is from $k=1$ to $k=n-1$ 
How do we know to use $k=2$ and to stop at $k=2$ for that first term?

$$7\sum_{k=1}^{n-1}\frac{1}{k}=7\left(1+\frac{1}{2}\right)+7\sum_{k=3}^{n-1}\frac{1}{k}$$
$$3\sum_{k=2}^{n}\frac{1}{k}=3\left(\frac{1}{2}+\frac{1}{n}\right)+3\sum_{k=3}^{n-1}\frac{1}{k}$$
$$4\sum_{k=3}^{n+1}\frac{1}{k}=4\left(\frac{1}{n}+\frac{1}{n+1}\right)+4\sum_{k=3}^{n-1}\frac{1}{k}$$
Do you see how to get the marked line now?
