My textbook has an example that says "Show that the sequence {${c_n}$} $= (-1)^n \frac{1}{n!} $ " converges, and find its limit.
It tells me that I must "find two convergent sequences that can be related to the given sequence" which the textbook states that the two possibilities are $a_n = \frac{-1}{2^n}$ and $b_n = \frac{1}{2^n}$.
My question is how did they find $a_n$ and $b_n$? Is there a way to find them algebraically since my text doesn't show or explain the process to do so?
The idea of the squeeze theorem is that you find two sequences, in your example $a_n$ and $b_n$, with whom you can bound the sequence $c_n$ (you are interested in), i.e., so that you can get for all $n>t$ ($t$ is some finite threshold) $$a_n\le c_n \le b_n.$$ If this holds, and you know that both $a_n$ and $b_n$ converge to the same limit $x$, then $c_n$ must converge to the same limit $x$ (after all, it is "sandwiched" from above and below by those two sequences; it has no choice but to converge, too). This is helpful if the limit of $c_n$ is hard to figure out, but it is easy to find sequences $a_n$ and $b_n$ that bound ("sandwich") $c_n$ and for which the limit can be found easily. This is also why the squeeze theorem is sometimes informally called "sandwich theorem".
In your example, $$c_n=(-1)^n\frac{1}{n!}<b_n=\frac{1}{2^n} \text{ for $n>3$},$$ because $n!=n\cdot (n-1)\cdot (n-2)\cdot \ldots > 2^n=2\cdot2\cdot\ldots\cdot 2$ and $\max\{(-1)^n\}=1$. Similarly, $a_n<c_n$ if $c_n$ is negative.
Now $a_n$ and $b_n$ are rather simple sequences, and their limit is $x=0$. The only difference is that $a_n<0$ converges from below and $b_n>0$ converges from above. By the squeeze theorem, we therefore know that $c_n\to 0$ as $n\to\infty$ as well.
How do you find those bounding sequences? This often happens on a case-by-case basis. But as a general rule, if your sequence $c_n$ consists of several terms, then try to bound each of them. Your $c_n$ is a product, so the first bound is $\max\{(-1)^n\}=1$ and the second is $\text{*something*}\ge \frac{1}{n!}$. In your textbook, $\text{*something*}=\frac{1}{2^n}\ge \frac{1}{n!}$, which holds for $n>3$ as shown above. Since both factors are positive, the product of two factors (of $b_n$), each of whom are weakly greater than the factors of another product ($c_n$), is also weakly larger. Hence $b_n\ge c_n$. Again, the important thing is that the sequence you find is simpler, so you just look at your sequence and think about how to simplify it, ideally so that is remains a close approximation for $n$ large.