Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$.

Question

Solution

Notice that $$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$ Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain $$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$ Thus, we obtain $$e>\frac{n}{\sqrt[n]{n!}}.\tag1$$ Moreover, we can find that, for $k=0,1,\cdots,n-1.$ $$\frac{n^k}{k!}< \frac{n^n}{n!}.$$ Thus \begin{align*} e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\ &< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\ &=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\ &=(2n+1)\cdot \frac{n^n}{n!}, \end{align*} which shows that $$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.\tag2$$ Combining $(1)$ and $(2)$, we have $$e>\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}\to e(n \to \infty).$$ By the squeeze theorem, $$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$

Please correct me if I'm wrong. Many thanks.

Answer 1

Using Riemann sums:

$$ \log\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^{n}\log\left(\tfrac{k}{n}\right)\to \int_{0}^{1}\log(x)\,dx = -1$$ hence $\frac{n}{\sqrt[n]{n!}}$ converges to $e$.

Answer 2

Seems okay another way could be $$\frac{n}{\sqrt[n]{n!}}=\exp({\ln n-\frac{\ln n!}{n}})=\exp\left(\frac{n\ln n-\ln n!}{n}\right)$$ ($\exp(x)=e^x$)

Then by continuity we have that $$\lim_{n\to\infty}\exp\left(\frac{n\ln n-\ln n!}{n}\right)=\exp\left(\lim_{n\to\infty}\frac{n\ln n-\ln n!}{n}\right)$$

Now you can find the $\lim$ by using Stolz theorem. $$\lim_{n\to\infty}\frac{n\ln n-\ln n!}{n}=\lim_{n\to\infty}(n+1)\ln(n+1)-\ln((n+1)!)-n\ln n+\ln n!=\lim_{n\to\infty}n\ln(n+1)-n\ln n=\lim_{n\to\infty}\ln(1+\frac1n)^n=1$$ Hence the whole $\lim$ equals to $e$

Answer 3

For the denominator, rewrite it as $ e^{\frac{\sum_{k=1}^{n} \log k}{n}}$ then approximate the sum of logs with the integral ($n \log n -n) $ to get the result.

Answer 4

You are right with your limit $e$. Using Stirling's formula you have putting $f(n)=\dfrac{n}{\sqrt[n]{n!}}$, $$\log f(n)=\log n-\frac{n\log n-n+O(\log n)}{n}\Rightarrow \log f(n)-1=\frac{O(\log n)}{n}$$ By definition of Big $O$ notation there is some positive constant $C$ and some $n_0$ such thar for all $n\gt n_0$ we have $$|\log f(n)-1|\le \frac{C\log n}{n}$$ Then $\log f(n)-1$ tends to $0$ so $f(n)$ tends to $e$ as you have show.