1. ## Question about the root test for series

Please see the attached worked problem.

Root Test example.pdf

My question is about the final limit:

$\lim_{n\rightarrow\infty} [\frac{n}{n} \cdot \frac{(n-1)}{n} \cdot (n-2)!]$

Can someone show me the steps to get to this point? I am confused as to how the limit becomes this. I know that it is because of the factorial, but it this from canceling due to a telescoping series?

Thanks!

2. So, by simple exponent rules: $a_n = \frac{(n!)^n}{n^{2n}} = \frac{(n!)^n}{\left(n^{2}\right)^n} = \left(\frac{n!}{n^2}\right)^n$

Now, to use the root test we consider: $\lim_{n \to \infty} \sqrt[n]{\left|a_n\right|} = \lim_{n \to \infty} \sqrt[n]{\left(\frac{n!}{n^2}\right)^n} = \lim_{n \to \infty} \frac{n!}{n^2}$

Recall that: $n! \ \ = \ \ n(n-1)! \ \ = \ \ n(n-1)(n-2)! \ \ = \ \ n(n-1)(n-2)\cdots 2 \cdot 1$

So, our limit becomes: $\lim_{n \to \infty} \frac{n(n-1)(n-2)!}{n \cdot n} = \lim_{n \to \infty} \left( \frac{n}{n} \cdot \frac{n-1}{n} \cdot (n-2)!\right)$ (just breaking fraction apart to get rid of the $n^2$)

Now, $\frac{n}{n}$ is simply 1 and $\lim_{n \to \infty} \frac{n-1}{n} = 1$.

So our limit finally becomes: $\lim_{n \to \infty} (n-2)! = \infty \ \ {\color{red}>} \ \ 1$