Finding the limit of $\frac {n}{\sqrt[n]{n!}}$

I'm trying to find $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$

I tried couple of methods: Stolz, Squeeze, D'Alambert

Thanks!

Edit: I can't use Stirling.

Solution 1:

Let $\displaystyle{a_n=\frac{n^n}{n!}}$. Then the power series $\displaystyle{\sum_{n=1}^\infty a_n x^n}$ has radius of convergence $R$ satisfying $\displaystyle{\frac{1}{R}=\lim_{n\to \infty} \sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}}$, provided these limits exist. The first limit is what you're looking for, and the second limit is $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$.

Added: I just happened upon a good reference for the equality of limits above, which gives a more general result which is proved directly without reference to power series. Theorem 3.37 of Rudin's Principles of mathematical analysis, 3rd Ed., says:

For any sequence $\{c_n\}$ of positive numbers, $$\liminf_{n\to\infty}\frac{c_{n+1}}{c_n}\leq\liminf_{n\to\infty}\sqrt[n]{c_n},$$ $$\limsup_{n\to\infty}\sqrt[n]{c_n}\leq\limsup_{n\to\infty}\frac{c_{n+1}}{c_n}.$$

In the present context, this shows that $$\liminf_{n\to\infty}\left(1+\frac{1}{n}\right)^n\leq\liminf_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$$ Assuming you know what $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$ is, this shows both that the limit in question exists (in case you didn't already know by other means) and what it is.

From the comments: User9176 has pointed out that the case of the theorem above where $\displaystyle{\lim_{n\to\infty}\frac{c_{n+1}}{c_n}}$ exists follows from the Stolz–Cesàro theorem applied to finding the limit of $\displaystyle{\frac{\ln(c_n)}{n}}$. Explicitly, $$\lim_{n\to\infty}\ln(\sqrt[n]{c_n})=\lim_{n\to\infty}\frac{\ln(c_n)}{n}=\lim_{n\to\infty}\frac{\ln(c_{n+1})-\ln(c_n)}{(n+1)-n}=\lim_{n\to\infty}\ln\left(\frac{c_{n+1}}{c_n}\right),$$ provided the latter limit exists, where the second equality is by the Stolz–Cesàro theorem.

Solution 2:

This is going to be a bit difficult (since apparently lots of things aren't allowed). Here's how I would do it (this is far from a complete solution but just a couple of hints):

I hope you know that $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n}$ (this is often taken as the definition of $e$).

You can show easily that the sequence $c_{k} = \left(1 + \frac{1}{k}\right)^k$ is monotonically increasing and that the sequence $d_{k} = \left(1 + \frac{1}{k}\right)^{k+1}$ is monotonically decreasing. This gives the squeezing $$\displaystyle \left(1 + \frac{1}{k}\right)^k = c_k \lt e \lt d_k = \left(1 + \frac{1}{k}\right)^{k+1}.$$

By taking the products $c_{1} c_{2} \cdots c_{n}$ and $d_{1} d_{2} \cdots d_{n}$ you can then show $$\displaystyle \frac{(n+1)^n}{n!} \lt e^n \lt \frac{(n+1)^{n+1}}{n!} $$ using a few manipulations.

Now extract roots on both sides of the last inequalities and you're there.

Solution 3:

By applying Cauchy-d'Alembert criterion we get that:

$$\lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\frac{1}{n}} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}} = \lim_{n\to\infty} \frac{(n+1)^{(n+1)}}{(n+1)!}\cdot \frac{n!}{n^n} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}= \lim_{n\to\infty} \frac{(n+1)^n}{n^n} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty} {\left(1+\frac{1}{n}\right)^{n}}=e. $$

Q.E.D.

Solution 4:

If $f(n)=\frac{n}{\sqrt[n]{n!}}$ and $g(n) = f(n)^n$ then

$$g(n) = \frac{n^n}{n!}$$

and taking the ratio of terms, removing the factorials and using $\frac{n+1}{n} = 1+\frac{1}{n}$,

$$ \frac{g(n+1)}{g(n)} = \left(1 + \frac{1}{n}\right)^n $$

You may recognise this as having a limit of $e$. It implies

$$\lim_{n \to \infty} \frac{g(n+1)}{g(n)} \frac{1}{e} = 1$$

and so multiplying a string of these together

$$\lim_{n \to \infty} \frac{g(n)}{e^n h(n)} = 1$$

for some function $h(n)$ which grows more slowly than $e^n$ or decays more gently than $e^{-n}$, [not that it matters, but $h(n)$ is about $1/\sqrt{2 \pi n}$] so taking the $n$-th root

$$\lim_{n \to \infty} \frac{f(n)}{e} = \lim_{n \to \infty} h(n)^{1/n} = 1$$

and so $\lim_{n \to \infty} f(n) = e$.

Solution 5:

If you take the log, it is:

$$\frac{1}{n}\sum_{k=1}^n \log\left(\frac{k}{n}\right)$$

Which is a Riemann sum for $\int_{0}^1 \log x$.

The indefinite integral is $F(x)=x\log x-x$ and $\lim_{x\to 0} x\log x -x =0$, and $F(1)=-1$.

You have to deal with the fact that this integral is an improper integral, but it "just works."