Using the Squeeze Theorem in Sequences 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?
 A: 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.
