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

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 $$e^{-1}$$.

• Because $e^x$ is continuous, and $e^{a_n/b_n}$ is our sequence – Jakobian Sep 27 '18 at 21:38

## 2 Answers

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: $$\lim_{n\to\infty} x_n = \exp(-1).$$

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

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

• This is what I used for deriving the result at first. – DreaDk Sep 27 '18 at 22:38
• @DreaDk Indeed it is a very effective method for that! – gimusi Sep 27 '18 at 22:42
• +1 for my favorite technique. This is the multiplicative case of Cesaro-Stolz. If $a_{n+1}/a_n\to L$ then $b_{n+1}-b_n\to \log L$ where $b_n=\log a_n$. By Cesaro-Stolz $b_n/n\to \log L$ and exponentiating we get $a_{n} ^{1/n}\to L$. – Paramanand Singh Sep 28 '18 at 5:44
• @ParamanandSingh I always forget to recall that fact! Thanks – gimusi Sep 28 '18 at 6:40