How to take the Limits of Logs

Question

How would you take the limit of

$$\frac{\log(n!)}{\log(n^n)}$$

as $n\rightarrow\infty$.

I believe you have to remove the log raising it to their base. Is this correct ?

Thanks.

Answer 1

Hint:

Use Sterling's approximation of $n!$:

$$n!\sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}$$

as $n\rightarrow\infty$.

Solution:

By Sterling's approximation:

$$\lim_{n\rightarrow\infty}\frac{\log(n!)}{\log(n^n)}=\lim_{n\rightarrow\infty} \frac{\log((n/e)^n\sqrt{2\pi n})}{\log n^n}$$

Use logarithm laws to see

$$\log((n/e)^n\sqrt{2\pi n})=n\log n-n+\log(\sqrt{2\pi n})$$

and $\log(n^n)=n\log(n)$.

Then

$$\frac{n\log n-n+\log(\sqrt{2\pi n})}{n\log n}=1-\frac{1}{\log n}+\frac{\log(\sqrt{2\pi n})}{n\log n}$$

The last two terms all go to $0$ as $n\rightarrow\infty$. Thus we conclude

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

Answer 2

Hints:

1)

$$ \ln(n!) = \sum_{k=1}^{n}\ln(k) \sim \int_{1}^{n}\ln(x)dx = n\ln n - n +1. $$

2)

$$ \ln(n^n) = n\ln n. $$