Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ when $n \rightarrow \infty$?

Question

Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ ? How to solve such type of questions considering $n \rightarrow \infty$?

Answer 1

Assume $n=k^2$ is a square. $\sqrt{n!}$ is the square root of a product of $n$ integers. So you can see it as a product of $n$ square roots of integers: $$\sqrt{\color{blue}{1}}\sqrt{2}\sqrt{3}\sqrt{\color{blue}{4}}\cdots{\sqrt{\color{blue}{k^2}}}$$ But $(\sqrt{n})!=k!$ is a product of $k$ integers: $$(\color{blue}{1})(\color{blue}{2})\cdots (\color{blue}{k})$$ Each of the factors in this last expression appear somewhere in the first product. Therefore the first product is much larger.

Answer 2

As alluded to in the comments, $(\sqrt{n})!$ doesn't make sense, so I'm going to compare the growth of $n!$ to the growth of $\sqrt{(n^2)!}$. Or, equivalently, compare $(n!)^2$ to $(n^2)!$.

Let $a_n = \frac{(n!)^2}{(n^2)!}$. Then \begin{align*} \frac{a_{n+1}}{a_n} &= \frac{((n+1)!)^2 \div (n!)^2}{((n+1)^2)! \div(n^2)!} \\ &= \frac{(n+1)^2}{(n + 1)^2 ((n + 1)^2 - 1)((n + 1)^2 - 2) \ldots(n+1)} \\ &= \frac{1}{((n + 1)^2 - 1)((n + 1)^2 - 2) \ldots(n+1)}. \end{align*}

So, the ratio between $(n!)^2$ and $(n^2)!$ very quickly approaches $0$, telling you that $(n^2)!$ grows much faster than $(n!)^2$.

Answer 3

Let $n = m^2$.

$a = \ln (\sqrt n)! = \ln 1 + \ln 2 + \cdots + \ln m$

$b = \ln \sqrt {n!} = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m + \ln (m+1) + \ln (m+2) + \cdots + \ln m^2\}$

$a - b = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m\} - \frac{1}{2} \{\ln (m+1) + \cdots + \ln (m+m) + \cdots + \ln m^2\}$

$a - b < \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m\} - \frac{1}{2} \{\ln 1 + \ln 2 + \cdots + \ln m\} = 0$

Hence $\sqrt n! < \sqrt {n!}$.

Answer 4

Assuming $n=k^2$ and using the Stirling's approximation: $$\lim_{n\to \infty} \frac{\sqrt {n!}}{(\sqrt {n})!}=\lim_{k\to \infty} \frac{\sqrt {k^2!}}{(\sqrt {k^2})!}=\lim_{k\to \infty} \sqrt{\frac{k^2!}{(k!)^2}}=\sqrt{\lim_{k\to \infty} \frac{k^2!}{(k!)^2}}=\\ \sqrt{\lim_{k\to \infty} \frac{\left(\frac{k^2}{e}\right)^{k^2}\cdot \sqrt{2\pi k^2}}{\left[\left(\frac{k}{e}\right)^{k}\cdot \sqrt{2\pi k}\right]^2}}= \sqrt{\lim_{k\to \infty} \frac{\left(k^2\right)^{k^2-k}}{e^{k^2-2k}\cdot \sqrt{2\pi}}}=\\ \sqrt{\frac{1}{\sqrt{2\pi}}\lim_{k\to \infty} \left(\frac{k^2}{e}\right)^{k^2-k}\cdot e^k}=\infty.$$ It implies that $\sqrt{n!}$ grows much faster than $(\sqrt{n})!$.