How can we prove that $L = \lim_{n \to \infty}\frac{\log\left(\frac{n^n}{n!}\right)}{n} = 1$

Question

How can we prove that $L = \lim_{n \to \infty}\frac{\log\left(\frac{n^n}{n!}\right)}{n} = 1$

This is part of a much bigger question however I have reduced my answer to this, I have to determine the limit of $\log(n^{n}/n!)/n$ when $n$ goes to infinity.

Apparently the answer is 1 by wolfram alpha but I have no clue how to get it. Any idea how I could proceed (without sirlings approximation as well).

Answer 1

Using an integral approximation of the sum of the logarithms of the first $n$ natural numbers gives:

$$n \log n - n=\int_{1}^n\log(t)dt\leq\underbrace{\sum_{k=1}^n\text{log}(n)}_{\log(n!)}\leq \int_{1}^{n+1}\text{log}(t)dt=\underbrace{(n+1)\log(n+1)-(n+1)}_{n\log(n)-n-1+\log(n)+(n+1)\log(1+1/n)}\\\implies n\log(n)-n\leq \log(n!)\leq n\log(n)-n+\underbrace{\left[-1+\log(n)+(n+1)\log(1+1/n)\right]}_{\alpha(n)}\\\implies n\log(n)-n\leq \log(n!)\leq n\log(n)-n+\alpha(n)\\\implies -n\log(n)+n\geq -\log(n!)\geq -n\log(n)+n-\alpha(n)\\\implies -n\log(n)+n-\alpha(n)\leq -\log(n!)\leq -n\log(n)+n\\\implies n-\alpha(n)\leq\underbrace{n\log(n)-\log(n!)}_{\log(\frac{n^n}{n!})}\leq n\implies 1-\frac{\alpha(n)}{n}\leq \frac{\log(\frac{n^n}{n!})}{n}\leq n$$

However we have that: $$\small -\frac{3\log(n)}{n}\leq -\frac{\log(n)}{n}-\underbrace{2\log(e^{1/n})}_{\large{\frac{2}{n}}}\leq-\frac{\log(n)}{n}-2\log(1+1/n)\leq\underbrace{\frac{1}{n}-\frac{\log(n)}{n}-\frac{n+1}{n}\log(1+1/n)}_{{\large -\frac{\alpha(n)}{n}}}\\\implies -\frac{3\log(n)}{n}\leq -\frac{\alpha(n)}{n}\implies 1-\frac{3\log(n)}{n}\leq1-\frac{\alpha(n)}{n}$$

Which by our first inequality gives us:

$$1-\frac{3\log(n)}{n}\leq \frac{\log(\frac{n^n}{n!})}{n}\leq 1\implies \left|\frac{\log(\frac{n^n}{n!})}{n}-1\right|\leq \frac{3\log(n)}{n}\\\implies \forall \epsilon >0\left(n>\frac{9}{\epsilon^2}\implies \left|\frac{\log(\frac{n^n}{n!})}{n}-1\right|\leq \frac{3\log(n)}{n}<\frac{3}{n^{1/2}}\leq \frac{3}{(9{\epsilon}^{-2})^{1/2}}=\epsilon\right)\\\implies \forall \epsilon>0\exists N\in \mathbb{N}:\forall n\in \mathbb{N}\left(n\geq N\implies \left|\frac{\log(\frac{n^n}{n!})}{n}-1\right|<\epsilon\right)\implies \lim_{n\to\infty}\frac{\log(\frac{n^n}{n!})}{n}=1$$

As required.

Answer 2

Applying Stirling's formula $$ n! \underset{(+\infty)}{\sim} \sqrt{2n \pi}\left(\frac{n}{e}\right)^n $$ Hence $$ \frac{n^n}{n!}\underset{(+\infty)}{\sim} \frac{n^n}{\sqrt{2n \pi}\left(\frac{n}{e}\right)^n}=\frac{e^n}{\sqrt{2n \pi}} $$ Then $$ \ln\left(\frac{n^n}{n!}\right)\underset{(+\infty)}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right) $$ So with $ln(e)=1$ $$ \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$$

here's your result

Answer 3

$$n!\sim_\infty\sqrt{2\pi n\vphantom{h}}\,\Bigl(\frac ne\Bigr)^{\!n},\enspace\text{hence }\quad \frac{n^n}{n!} \sim_\infty\frac{\mathrm e^n}{\sqrt{2\pi n\vphantom{h}}} $$ so $$\frac1n\,\log\biggl(\frac{n^n}{n!}\biggr)\sim_\infty\frac1n\Bigl(\frac12\log(2\pi n)+n\Bigr)=\underbrace{\frac{\log(2\pi n)}{2n}}_{\substack{\downarrow\\0}} +1.$$

Answer 4

By Stolz-Cesaro

$$\lim_{n \to \infty}\frac{\log\left(\frac{n^n}{n!}\right)}{n} = \lim_{n \to \infty}\frac{\log\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)-\log\left(\frac{n^n}{n!}\right)}{n+1-n} = \lim_{n \to \infty}\log\left(\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}\right)=\\= \lim_{n \to \infty}\log\left(1+\frac1n\right)^n=1$$