I have to find the limit of this problem. I have found that with problems like this, I always get wrong answers when I try to use Cauchy's second limit theorem. I do not understand how to solve this problem. Please help me.
(This is a general problem not specific problem.)
Cauchy's second theorem on limits says that for a sequence $\{a_n\}$ of positive values, if $$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n} = \ell,$$ then $$\lim_{n\rightarrow \infty} (a_n)^{1/n} = \ell.$$
You are wanting the limit $\displaystyle \lim_{n\rightarrow \infty} (a_n)^{1/n}$ with $a_n = \left(\frac{n!}{n^n}\right)^n$. What happens when we simplify the ratio $\frac{a_{n+1}}{a_n}$? \begin{eqnarray} \frac{a_{n+1}}{a_n} &=& \frac{\left(\frac{(n+1)!}{(n+1)^{(n+1)}}\right)^{n+1}}{\left(\frac{n!}{n^n}\right)^n}\\ &=& \frac{\left(\frac{n!}{(n+1)^{n}}\right)^{n+1}}{\left(\frac{n!}{n^n}\right)^n}\\ &=& \left(\frac{n!}{(n+1)^{n}}\right)\cdot \left[\frac{\left(\frac{n!}{(n+1)^{n}}\right)}{\left(\frac{n!}{n^n}\right)} \right]^n\\ &=& \left(\frac{n!}{(n+1)^{n}}\right)\cdot \left[\left(\frac{n}{n+1}\right)^n \right]^n.\\ \end{eqnarray} The first factor in the product, $\left(\frac{n!}{(n+1)^{n}}\right)$, is bounded above by $1$. The second factor, $\left(\frac{n}{n+1}\right)^{n^2}$, approaches $0$ in the limit (hint: use L'Hôpital's rule to show $n^2\log(n/(n+1)) \rightarrow -\infty$). Therefore, $$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n} = 0.$$