What is the rule for manipulating the boundaries of a summation?
Summations are not like integrals. When you calculate an integral, swapping the boundaries does change the sign of the result. However, when talking about summations:

$$ \sum_{n=-\infty}^{-1}a(n)=\sum_{n=-1}^{-\infty}a(n)$$

This holds because it is the same if we perform the sum of $a(-1)+a(-2)+...+a(-k)$ for any $k&gt;0$ in any order. Namely, you can order those terms as you wish as long as all the terms are present. For infinite summations, this is true if the series is absolutely convergent.

In your example, they just made the change of variables $m=-n-1$ to get the summation from $0$ to $+\infty$. It may be confusing because they kept the original name of the index ($n$) instad of replacing it with the new one ($m$), but remember that they are just indexes or dummy variables, so in the end it doesn't matter. You can check that if $m=-n-1=-(n+1)$, then the boundaries change:

$$n=-1 \to m=-(-1+1)=0$$
$$n=-\infty \to m=-(-\infty)=\infty$$

where I used the fact that a finite number is negligible against infinity. So the summation can be written as:

$$\begin{align} -\sum_{n=-\infty}^{-1} \alpha^nz^{-n}&amp;=-\sum_{m=\infty}^{0} \alpha^{-(m+1)}z^{-(-(m+1))}\\
&amp;=-\sum_{m=0}^{\infty} \alpha^{-(m+1)}z^{-(-(m+1))}\\
&amp;=-\sum_{m=0}^{\infty} \alpha^{-(m+1)}z^{m+1}\\
&amp;=-\sum_{m=0}^{\infty} (\alpha^{-1}z)^{m+1}
\end{align}$$

And that's the same result you wrote in your question.