TITLE: evaluating $\lim_{n \to \infty }(\frac{n!}{n^n})^\frac{2n^4 + 1}{5n^5 + 1}$
QUESTION [5 upvotes]: I was trying to work out the following limit 
$$\lim_{n \to \infty }\Big(\frac{n!}{n^n}\Big)^{\Large\frac{2n^4+1}{5n^5+1}}$$
I have some idea about how to do it. Basically I expanded the $n!$ and divided each term with n to get 
$$\lim_{n \to \infty }\Big(\prod_{r=0}^{n-1}\Big(\frac{n-r}{n}\Big)\Big)^{\Large\frac{2n^4+1}{5n^5+1}}$$
and If I take logarithm then I can converte the summation into an integral but I have no clue how to go about treating the $\dfrac{2n^4+1}{5n^5+1}$ part which appears as coefficient in the summation.

REPLY [4 votes]: We can write the given sequence $a_{n}$ as $$a_{n} = \left\{\left(\frac{n!}{n^{n}}\right)^{1/n}\right\}^{(2 + 1/n^{4})/(5 + 1/n^{5})} = \{b_{n}^{1/n}\}^{c_{n}}\tag{1}$$ where $$b_{n} = \frac{n!}{n^{n}}, \, c_{n} = \frac{2 + (1/n^{4})}{5 + (1/n^{5})}$$ Now we can see that $c_{n} \to 2/5$ as $n \to \infty$ and $$\frac{b_{n + 1}}{b_{n}} = \frac{(n + 1)!}{(n + 1)^{n + 1}}\cdot\frac{n^{n}}{n!} = \dfrac{1}{\left(1 + \dfrac{1}{n}\right)^{n}} \to \frac{1}{e}\text{ as }n \to \infty$$ Hence $b_{n}^{1/n} \to 1/e$. From $(1)$ it is now clear that $a_{n} \to (1/e)^{2/5} = e^{-2/5}$ as $n \to \infty$.
Here we have used the following standard theorem
If $a_{n}$ is a sequence of positive numbers and $a_{n + 1}/a_{n} \to L$ as $n \to \infty$ then $a_{n}^{1/n} \to L$ as $n \to \infty$.