What is the rule for manipulating the boundaries of a summation?

Question

When working with DTFTs or $\mathcal Z$-transforms, we sometime get summations that do not go from $n=0$ to $+\infty$. For example, suppose we have the sequence $x(n) = -\alpha^n u(-n-1)$. To find the $\mathcal Z$-transform: \begin{align} X(z)&=\sum_{n=-\infty}^\infty x(n)z^{-n}\\ &=-\sum_{n=-\infty}^{-1} \alpha^nz^{-n} \end{align}

I understand this, the step function(which was shifted and inverted) causes the sequence to exist for all values from $-\infty$ to $-1$, and is $0$ everywhere else. However, the next part of the solution confuses me: $$=-\sum_{n=0}^{\infty} (\alpha^{-1}z)^{n+1} $$ I am not sure how they managed to change the boundaries.

• From my interpretation of the result , a flip of the boundaries $\sum_{n=-\infty}^{-1}$ to $\sum_{n=-1}^{\infty}$ for any sequence will cause the sign of $n$ to change. Is this true?
• Then, to get from $\sum_{n=-1}^{\infty}$ to $\sum_{n=0}^{\infty}$,do you just need to add $1$ to wherever here is an $n$?
• Should you always first flip the boundary (when necessary) and then only change the index afterwards?

Answer 1

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 $k>0$ 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:

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

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