limit of $\frac{\log(n^n)}{\log((2n)!)}$ as $n$ approaches $\infty$

Question

I'm trying to solve this:

limit of $\frac{\log(n^n)}{\log ((2n)!)}$ as $n$ approaches $\infty$

I know $\log(n^n)=n\log(n)$ using logarithms properties but what about $\log((2n)!)$? Idk how to solve this monstrous denominator.

Any help?

Answer 1

We will use the Stolz-Cesaro Theorem and L'Hospital's Rule: $$\lim\limits_{n\to\infty}\frac{\log(n^n)}{\log(2n!)}=\lim\limits_{n\to\infty}\frac{\log\left[(n+1)^{n+1}\right]-\log\left(n^n \right)}{\log[(2n+2)!)-\log((2n)!]}=\lim\limits_{n\to\infty}\frac{\log\left[\left(1+\frac{1}{n}\right)^n \cdot (n+1)\right]}{\log[(2n+1)(2n+2)]}=\lim\limits_{n\to\infty}\frac{\log(n+1)}{\log(2n+1)+\log(2n+2)}=\lim\limits_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{2}{2n+1}+\frac{2}{2n+2}}=\lim\limits_{n\to\infty}\frac{(2n+1)(2n+2)}{(n+1)(8n+6)}=\frac{1}{2}.$$
Note that the way I applied L'Hosipital's Rule is not so rigorous because we are only allowed to differentiate functions. The proper way to write this would have been to compute $\lim\limits_{x\to\infty}\frac{\log(x+1)}{\log(2x+1)+\log(2x+2)}$ using L'H, take the sequence $x_n=n$ and use the definition of continuity in terms of sequences to conclude that $\lim\limits_{x\to\infty}\frac{\log(x+1)}{\log(2x+1)+\log(2x+2)}=\lim\limits_{n\to\infty}\frac{\log(n+1)}{\log(2n+1)+\log(2n+2)}=\frac{1}{2}$.
The reason why I decided to "differentiate sequences" through abuse of notation is that it looks more compact and recently even some books do this.

Answer 2

One possible approach might be as below.

Using the fact that for $n>1$, $n\log(n)>log(n!)$

$$\lim_{n\rightarrow \infty}\frac{\log(n^n)}{\log(2n!)}=\lim_{n\rightarrow \infty}\frac{n\log(n)}{\log(1)+\log(2)+....+\log(2n)}\geq \lim_{n\rightarrow \infty}\frac{n\log(n)}{2n\log(2n)}=\lim_{n\rightarrow \infty}\frac{\log(n)}{2\log(2n)}=\frac{1}{2}$$

Now the question is if you can show that for every $x<\frac{1}{2} $ there exist $n\in \mathbb{N}$ such that $\frac{\log(n^n)}{\log(2n!)}>x.$

Answer 3

According to Stirling's approximation, we have $\ln (n!) \approx n \ln n - n$. Thus, $$ \frac{\ln (2n!)}{\ln n^n} \approx \frac{2n \ln (2n) - 2n}{n \ln n} = 2\left[ 1 + \frac{\ln 2 - 2}{\ln n}\right] $$ The fraction inside the brackets goes to zero as $n \to \infty$, so the limit of the fraction I've written goes to 2, and $$ \lim_{n \to \infty} \frac{\ln n^n}{\ln (2n!)} = \frac{1}{2}. $$