Solving $\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$

Question

I was solving a convergence of a series and this limit popped up:

$$\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$$

I needed this limit to be $0$ and it is in fact (according to WolframAlpha), but I just don't see how to get the result.

Answer 1

Since $n! \sim \sqrt{2\pi n} (n/e)^n$ by Stirling's approximation, we have

$$\frac{n^n}{e^nn!} = \frac{(n/e)^n}{n!} \sim \frac{(n/e)^n}{\sqrt{2\pi n} (n/e)^n} = \frac{1}{\sqrt{2\pi n}} \to 0.$$

Answer 2

Almost as good: write the expression as $$ L = e^{n \log n - n - \log n!} = e^{n \log n -n -\sum_{k=1}^{n} \log k} $$ and use the bounds on the sum: $$ \int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k < \int_{1}^{n+1} \log x dx $$ to get the same result without Stirling.

Answer 3

The ratio of two consecutive values is $$ \left.\frac{(n+1)^{n+1}}{e^{n+1}(n+1)!}\middle/\frac{n^n}{e^nn!}\right.=\frac{\left(1+\frac1n\right)^n}{e}\tag{1} $$ Taking the log of $(1)$ gives $$ n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)-1 =-\frac1{2n}+O\left(\frac1{n^2}\right)\tag{2} $$ Since $$ \int_1^{n+1}\frac{\mathrm{d}x}{x}\le\sum_{k=1}^n\frac1k\le1+\int_1^n\frac{\mathrm{d}x}{x}\tag{3} $$ we have $$ \sum_{k=1}^n\frac1k=\log(n)+O(1)\tag{4} $$ Therefore, summing $(2)$ using $(4)$ yields $$ \log\left(\frac{n^n}{e^nn!}\right)=-\frac12\log(n)+O(1)\tag{5} $$ Thus, we have $$ \frac{n^n}{e^nn!}\le\frac{c}{\sqrt{n}}\tag{6} $$ The limit sought is therefore, $0$.

Answer 4

Consider that: $$\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) = n, $$ hence: $$ n! = \prod_{m=2}^{n} m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}$$ and: $$\frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k}.\tag{1}$$ Since the sequence defined by: $$ a_k = \left(1+\frac{1}{k}\right)^{k+\frac{1}{3}} $$ is increasing towards $e$, from $(1)$ it follows that: $$ \frac{n^{n+\frac{1}{3}}}{n!}=\prod_{k=1}^{n-1} a_k \leq e^{n-1}\tag{2} $$ hence: $$ \frac{n^n}{n!e^n}\leq\frac{1}{e\sqrt[3]{n}},\tag{3}$$ proving our claim.