Proving $\lim((n!)(\frac{e}{n})^n) = \infty$ using elementary method

Question

Is there some easy way to show that $\lim_{n\to\infty} ((n!)(\frac{e}{n})^n) = \infty$ as $n \to \infty$? It looks like Wolfram alpha is using some kind of expansion method to numerically compute this? Is there a more elementary method? Thanks a lot!

Answer 1

Take the log of the limit to get

$$\ln(L)=\lim_{n\to\infty}\ln(n!)-n\ln(n)+n$$

Note that we have by trapezoidal approximations of the integral of $\ln$:

$$\ln(n!)=\sum_{k=1}^n\ln(k)\ge\frac{\ln(n)+\ln(1)}2+\int_1^n\ln(x)~\mathrm dx=\frac12\ln(n)+n\ln(n)-n+1$$

Hence we have

$$\ln(L)\ge\lim_{n\to\infty}\frac12\ln(n)+1=\infty$$