$x_{1}[k]$ is a function with an argument $k$

$$f(z)=\sum_{k=-\infty }^{+\infty} \left( x_{1}[k] \left( \sum_{n=-\infty }^{+\infty} \frac{x_{2}[n-k]}{z ^{n} } \right) \right)$$


$$n:-\infty\rightarrow +\infty$$

$$ \left(\therefore ~~p:-\infty\rightarrow +\infty\right)$$

$$f(z)=\sum_{k=-\infty }^{+\infty} \left( x_{1}[k] \left( \sum_{p=-\infty }^{+\infty} \frac{x_{2}[p]}{z ^{p+k} } \right) \right)$$

What I can't get currently is the range of $p~$ stated above.

As endpoints of the range of $k$ are not infinite things ,I can get $\left(p:-\infty\rightarrow +\infty\right) ~\text{as for instance }~\left(k:-10^{9}\rightarrow +10^{9}\right)$

However the actual endpoints of range of $k$ are infinite values so as $k=\infty~$ ,


is held and as $n=-\infty$, I can get $p=-\infty-\infty=-\infty$ however as $n=\infty$,


So how do I interpret like as above formula arose?

I thought below.

As $k=\infty$ and $n=\infty$ , $~~~n-k$ seems takes 0 however $n$ has the priority(which means that the value of n was determined later the determination of value of k) so $\infty_{\text{k}}<\infty_{\text{n}}$ holds so actually the below equation can be held.


Is this thought correct?

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