# How do I deduce the range of variable which endpoints of it are infinite values of change of variable?

$$x_{1}[k]$$ 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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