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

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

$$p:=n-k$$

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

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

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

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 $k=\infty~$ ,

$$p=n-(\infty)$$

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

$p=\infty-\infty=0?$

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.

$$n-k=\infty_{n}-\infty_{k}=\infty$$

Is this thought correct?

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You are thinking about this totally wrong. Here $\infty$ is just a notation; summing as $n$ goes from $-\infty$ to $\infty$ just means that you sum over each integer value of $n$. Since all of this is happening inside the summation over $k$, $k$ is just a fixed constant as you are doing this. So, as $n$ takes each possible integer value, the quantity $p=n-k$ will also take each integer value (you're just shifting the index of when each integer value appears by $k$).

In general, you should not attempt to think about these things in terms of arithmetic rules about how the top and bottom numbers of the summation transform. Instead, think about exactly what the set of terms you are adding up is. In this case, for each integer $k$, we have an $n$th term for each integer $n$. For each fixed $k$, this is equivalent to having a term for each integer value of $n-k$.

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