Multiplying an infinite sequence This is a "challenge" problem in the current text I'm studying.
When simplified, the product,
$$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\cdots\left(1-\frac{1}{n}\right),$$ becomes which of the following?
I spent quite a bit of time on the problem a few weeks ago and could never get my answer to match any of the answers given.
 A: NOTE:  Didn't see AWertheim's comment when I posted this.  Not trying to steal answers >.<
$$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\cdots\left(1-\frac{1}{n}\right)=$$$$\left(\frac{2}{\color{blue}{3}}\right)\left(\frac{\color{blue}{3}}{\color{red}{4}}\right)\left(\frac{\color{red}{4}}{\color{green}{5}}\right)\left(\frac{\color{green}{5}}{\color{purple}{6}}\right)\cdots\left(\frac{\color{orange}{n-1}}{n}\right)$$
A: This is the same thing as writing:
$$(\frac{2}{3})(\frac{3}{4})(\frac{4}{5})...(\frac{n-1}{n})$$
Which can be written as:
$$\lim_{n \to\infty} \frac{\left(\frac{(n-1)!}{n!}\right)}{(\frac{1!}{2!})}$$
$$= \lim_{n \to\infty}\left(\frac{(n-1)!}{n!} \frac{2!}{1!}\right) = \lim_{n \to\infty}\left(\frac{(n-1)!}{n*(n-1)!} \frac{2}{1}\right) = \lim_{n \to\infty}\left(\frac{2}{n}\right) \to \frac{2}{\infty} \to 0$$
Another way to think about it is that when you multiply it out, every number gets canceled out except a $2$ in the numerator, and $n$ in the denominator (see Arthur's answer). Since $n$ approaches infinity, $\frac{2}{n}$ gets infinitely close to $0$. So we say that the limit as $n$ approaches infinity, of the sequence is equal to  $0$.
