2

I want to calculate the following limit:

$\lim\limits_{n \to \infty} \frac{log_2(n!)}{nlog_2(n)}$

I know that $\lim\limits_{n \to \infty} \frac{n!}{n^n}=0$, and I think that, since the logarithm is a monotonically increasing function with no upper bounds, the limit I want to calculate will be equal to 0. But I cannot prove it.

I would also like to know if what I assumed before is correct for any functions, and if so if there is a proof for it. That is:

If for 2 functions it is true that: $\lim\limits_{x \to \infty} \frac{f(x)}{g(x)}=0$

is it also true that $\lim\limits_{x \to \infty} \frac{h(f(x))}{h(g(x))}=0$

where h is a monotonically increasing function with no upper bounds?

michalis vazaios
  • 1,252

1 Answers1

1

Observe that you can split the numerator into $$\log{n}+\log{(n-1)}+...+\log{2}$$ So when dividing by $n\log{n}$, we can split the fraction into $$\frac{\log{n}}{n\log{n}}+\frac{\log{(n-1)}}{n\log{n}}+\frac{\log{2}}{n\log{n}}$$ Viewed this way, we can see the limit is indeed $0$.

For your second question, observe that $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$ means that for any (small) $\epsilon > 0$, there exists $x_{0}$ such that whenever $x>x_{0}$ we have $$\frac{f(x)}{g(x)}<\epsilon$$ We can rephrase this as "eventually $f(x)<\epsilon g(x)$". Now, if $h$ is strictly increasing, this implies that $$h(f(x)) < h(\epsilon g(x))$$ However, what we want to say is that $$h(f(x))<\epsilon h(g(x))$$ An general proof is escaping me (and perhaps it's not true in general), but I'll point out that if $h$ is assumed to be convex (or any function satisfying $h(ax)<ah(x))$, then this follows immediately.

pwerth
  • 3,880