How does $Θ(\log(n!))=Θ(\log(n^n)$?

I understand why $Θ(\log(n!))=Θ(n\log(n))$ and $Θ(\log(n^n))=Θ(n\log(n))$, therefore $Θ(\log(n!))=Θ(\log(n^n)$. But I am having trouble reconciling this with the following:

$$\lim_{x\to \infty}\left(\frac{\log_2(x^x)}{\log_2(x!)}\right)=\infty. $$

So how could there be constants such that $Θ(\log(n!))=Θ(\log(n^n))$?

Where has my understanding gone wrong?


2 Answers 2


As a matter of fact, $$\lim_{x\to \infty}\frac{\log_2(x^x)}{\log_2(x!)}=1.$$ So there is no problem to reconcile.

Looking at the first revision of the question, it seems to me that you are confused about the fact that $\frac{n^n}{n!}$ has different behaviour from $\frac{\log n^n}{\log n!}$: $$\lim_{n\to+\infty}\frac{n^n}{n!}=+\infty,\qquad\text{while}\qquad\lim_{n\to+\infty}\frac{\log(n^n)}{\log(n!)}=1<+\infty.$$ There is no problem with this. Unboundedness of ratios of functions is not preserved by applying arbitrary functions such as $\log$, because $\exp$ (the inverse of $\log$) is not polynomially bounded. For a simpler example, $$\lim_{n\to+\infty}\frac{n^2}{n}=+\infty,\qquad\text{while}\qquad\lim_{n\to+\infty}\frac{\log(n^2)}{\log(n)}=2<+\infty.$$

In general, a function $h\colon\mathbb R_{>0}\to\mathbb R_{>0}$ satifies the condition that $$\limsup_{n\to+\infty}\frac{f(n)}{g(n)}<+\infty\implies\limsup_{n\to+\infty}\frac{h(f(n))}{h(g(n))}<+\infty$$ for all $f,g\colon\mathbb N\to\mathbb R_{>0}$ if and only if there exists a constant $c$ such that $$y\le2x\implies h(y)\le ch(x)\tag1$$ for all $x,y\in\mathbb R_{>0}$. For example, (1) holds for all function of the form $h(x)=x^\alpha$ for some $\alpha$. On the other hand, any function $h$ satisfying (1) is polynomially bounded, specifically, $h(x)=O(x^{\lceil\log_2c\rceil})$.


$$\log (n!)=\log\left(\prod_{i=1}^{n}i\right)$$ $$=\log(1)+\log(2)+\dots+\log n$$ $$=\sum_{i=1}^{n}\log i\leq n\log n.$$ $$\sum_{i=1}^{n}\log i=\sum_{i=1}^{\frac{n}{2}}\log i+\sum_{i=\frac{n}{2}+1}^{n}\log i$$ $$\geq\frac{n}{2}\log(\frac{n}{2})=\Theta(n\log n).$$

Since $$\log n^n=n\log n$$ $$\implies \log (n!)=\Theta(\log n^n).$$


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