Which grows faster? $(n-2)!, n^3 , e^n, 3^n, \log(n!), (\sqrt{2})^{\log(n)}, n^{1/\log(n)}$

Question

I think that the order is $\sqrt{2}^{\log(n)},n^3,e^n,3^n,(n-2)!$...But I have two question the first one does $\log(n!)$ is approximately equal to $n^n$? if yes then it grows the fastest. The other question is how can I know if $n^{1/\log(n)}$ is smaller or bigger than $e^n$, $3^n$, and $\sqrt{2}^{\log(n)}$? (I know it is exponential so to put it in its order between exponential)

Answer 1

$$(\sqrt{2})^{\ln(n)} = 2^{\frac{1}{2}\ln(n)} = 2^{\ln(\sqrt{n})}= 2^\frac{log_2(\sqrt{n})}{\log_2(e)} = n^\frac{1}{2\log_2(e)}$$

Since $\frac{1}{2\log_2(e)} < 3$, because $1 < 6\log_2(e)$, we have that $(\sqrt{2})^{\ln(n)}$ grows slower than $n^3$.

What's more, obviously $e^n$ grows slower than $3^n$ which then grows slower than $(n-2)!$, since it is easy to prove (by induction) that $n! > (\frac{n}{3})^n$ starting from some big enough $n$. So for a moment we get:

$(\sqrt{2})^{\ln(n)}$ then $n^3$ then $e^n$ then $3^n$ then $(n-2)!$

We're left with only $\ln(n!)$ and $n^\frac{1}{\ln(n)}$

Looking at the latter $n ^ \frac{1}{\ln(n)} = \exp(\frac{1}{\ln(n)} \cdot \ln(n)) = e $, so it's just a constant, hence it'll take our first honored place.

What's more, since $n! < n^n$ (which is obvious or again - easy provable by induction), we get $\ln(n!) < n\ln(n) < n^3$

So there's only 2 possibilities for $\ln(n!)$, either between $e$ and $(\sqrt{2})^{\ln(n)}$ or $(\sqrt{2})^{\ln(n)}$ and $n^3$.

But from what we stated earlier $n! > (\frac{n}{3})^n$ and hence $\ln(n!) > n\ln(\frac{n}{3})$.

And since $n > n^\frac{1}{2\log_2(e)} = (\sqrt{2})^{\ln(n)}$, we get:

$n^\frac{1}{\ln(n)}$ then $(\sqrt{2})^{\ln(n)}$ then $\ln(n!)$ then $n^3$ then $e^n$ then $3^n$ then $(n-2)!$