Limit of $\frac{\sqrt[n]{n!}}{n}$ using Stolz-Cesáro

Question

I'm trying to calculate the following limit using Stolz-Cesáro

$$\lim_{n \to \infty}{\frac{\sqrt[n]{n!}}{n}} \qquad {(1)}$$

I know the result is $e^{-1}$ using a different method. In the book I'm using it's suggested to use the sequences $a_n=\ln(\frac{n!}{n^n})$ and $b_n=n$.

In this case we have :

$$\lim_{n \to \infty}{\frac{\ln \left(\frac{n!}{n^n}\right)}{n}}=\lim_{n \to \infty}\left[{ \frac{\ln \left(\frac{n!}{n^n}\right)-\ln \left(\frac{(n+1)!}{(n+1)^{n+1}}\right)}{n+1-n}}\right]=\lim_{n \to \infty}{\ln \left (\frac{n^n}{(n+1)^n}\right)}=-1$$

Not sure how this implies that $(1)$ is equal to $e^{-1}$.

Answer 1

Due to continuity of exponential function: $$ \lim_{n\to\infty} x_n = \lim_{n\to\infty} \exp(\log(x_n)) = \exp\left(\lim_{n\to\infty} \log(x_n)\right). $$ In your case $x_n={\frac{\sqrt[n]{n!}}{n}}$ and you have already proved that $$\lim_{n\to\infty} \log(x_n) = -1.$$ Therefore: $ \lim_{n\to\infty} x_n = \exp(-1). $

Answer 2

As an alternative by ratio-root criteria with

$$a_n =\frac{\sqrt[n]{n!}}n= \sqrt[n]{\frac{n!}{n^n}} \quad b_n=\frac{n!}{n^n}$$

we have that

$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$

thus since

$$\frac{b_{n+1}}{b_n}=\frac{{(n+1)!}}{n!}\frac{n^n}{(n+1)^{n+1}}=\frac{1}{\left(1+\frac1n\right)}\to \frac1e \implies a_n = \frac{\sqrt[n]{n!}}n\to \frac1e $$