What is bigger $f(n) = \sum_{i=1}^{n^3} \log i, g(n) = \log((\log\sqrt{n})!)$?

Question

What is bigger $f(n) = \sum_{i=1}^{n^3} \log i, g(n) = \log((\log\sqrt{n})!)$?

So I believe it is $f(n)$, I want to prove it goes to infinity faster than $g(n)$.

So: $f(n) = \sum_{i=1}^{n^3} \log i = \log(1 \cdot 2 \cdot ... \cdot n^3) = \log(n^3!)$

And from now it is "easy" for me to see that $f(n)$ goes to infinity faster than $g(n)$, How can I show that by this: $\lim \frac{f(n)}{g(n)} = \infty$?

Answer 1

Since $k! \le k^k$ (each factor is at most $k$), we have $$ g(n) \le \log \big( (\log\sqrt n)^{\log\sqrt n} \big) = (\log\sqrt n)(\log\log\sqrt n) < (\log n)(\log \log n). $$ On the other hand, $$ f(n) \ge \sum_{i=n}^{n^3} \log i \ge (\log n) \sum_{i=n}^{n^3} 1 > (\log n)(n^3-n). $$ Therefore $$ \frac{f(n)}{g(n)} >\frac{n^3-n}{\log\log n}, $$ which tends to infinity.

Answer 2

From the Stirling's approximation we have $$g\left(n\right)=\log\left(\left(\log\left(\sqrt{n}\right)\right)!\right)=\log\left(\Gamma\left(\log\left(\sqrt{n}\right)+1\right)\right)\sim\left(\log\left(\sqrt{n}\right)+1\right)\log\left(\log\left(\sqrt{n}\right)+1\right) $$ and, by Abel's summation, it is not difficult to prove that $$\sum_{k\leq x}\log\left(k\right)=x\log\left(x\right)-x+O\left(1\right) $$ so $$f\left(n\right)=\sum_{k\leq n^{3}}\log\left(k\right)=3n^{3}\log\left(n\right)-n^{3}+O\left(1\right)\sim3n^{3}\log\left(n\right) $$ hence $$\lim_{n\rightarrow\infty}\frac{f\left(n\right)}{g\left(n\right)}=\lim_{n\rightarrow\infty}\frac{3n^{3}\log\left(n\right)}{\left(\log\left(\sqrt{n}\right)+1\right)\log\left(\log\left(\sqrt{n}\right)+1\right)}=\infty$$ as wanted.