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Question 3.1 d from Chapter 3 of Oppenheim's Signals & Systems I have

$$x[n]=(-1)^n{u[-n]-u[-n-8]}$$ $$h[n]=u[n]-u[n-8]$$

and the question asks for $y[n]$ which is the convolution of $x[n]$ and $h[n]$ namely

$$y[n]=x[n]\star h[n]$$

I'm stuck with the limits on the sum

$$\sum_{k=-\infty}^{\infty}{\left((-1)^{k}(u[-k]-u[-k-8])\right)\left(u[n-k]-u[(n-8)-k]\right)}$$

Should the limit be '$0$' or '$n-k$' or '$n-8$'?

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    $\begingroup$ The limits are always the same: $-\infty$ to $\infty$. However, in many instances, the signals have value $0$ for infinitely many arguments, e.g. for all $k < 0$ or all $k > 35$, say, in which case, instead of wasting time adding infinitely many $0$s in the sum, you can reduce the limits on the sum and add only those numbers which might possibly be nonzero .nonzero $\endgroup$ – Dilip Sarwate Nov 16 '13 at 14:17
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Here is another example of how incredibly poor choice of notation misleads students everywhere. I do not possess a copy of any of the various tomes on signal processing that seem to be revered as the fifth Gospel on this site, and so I do not know if the book cited by the OP actually has written $(1)$, but I assert that it is against mathematical common sense to write things like $$y[n] = x[n] \star h[n].\tag{1}$$ It is generally accepted mathematical convention that is a symbol has the same meaning everywhere it appears in an equation or expression. For example, when we write $$y(t) = x(t)h(t),~ -\infty < t < \infty, \tag{2}$$ we mean by this that for every choice of real number $t$, the value of $y(t)$ is the same as the product of the values of $x(t)$ and $h(t)$; for example, $y(3) = x(3)h(3)$, and $y(313.012) = x(313.012)h(313.012)$, ans so on and so forth. So how are we supposed to make sense of this monstrosity $(1)$? Clearly $y[3]$ is not $x[3]\star h[3]$ whatever meaning we might ascribe to the latter quantity; the value of $y[3]$ depends on a lot of other values of $x$ and $h$. The poor notation misleads students into writing things like

$$\sum_{k=-\infty}^{\infty}{\left((-1)^{k}(u[-k]-u[-k-8])\right)\left(u[-n-k]-u[(-n-8)-k]\right)}$$

which is not the correct expression for the convolution (or the correlation, for that matter) of $$x[n]=(-1)^n{u[-n]-u[-n-8]}\quad \text{and} \quad h[n]=u[n]-u[n-8]\tag{3}$$

The correct way to do this is write $$y[n] = (x\star h)[n] = \sum_{k=-\infty}^\infty x[k]h[n-k]$$ and then substitute $n-k$ for $n$ in $h[n]=u[n]-u[n-8]$ to get $h[n-k] = u[n-k]- u[n-k-8]$ and not $u[-n-k]-u[(-n-8)-k]$ the way that the OP has it.

Some initial thought might have shown that (assuming that $u[n]$ is the unit step), that $h[n]$ is $0$ for $n < 0$ (since then both $u[n]$ and $u[n-8]$ are $0$, and that $h[n] = 0$ for $n \geq 8$ (since then both $u[n]$ and $u[n-8]$ are $1$. Consequently for any given fixed integer $n$, $h[n-k]$ is nonzero only for $$0 \leq n-k \leq 7 \Rightarrow n-7 \leq k \leq n$$ and so the limits on the sum could be reduced from $-\infty$ and $\infty$ to $n-7$ and $n$ (cf. my comment on the OP's question). I will leave it to the OP to figure out whether the fact that $x[n]$ is also time-limited can be used to further reduce the limits on the sum. Hint: the answer to this might depend on the specific value of $n$, and you may need to try several different values of $n$, e.g. $0, 3, 8, 10, -40$ etc to see if you can discern a pattern.

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