Roots of Factorials

We want to determine which mathematical relation ( $>, < , =$ ) that satisfy the question mark (?) in

$\large \sqrt[9]{9!} \quad ? \quad \sqrt[10]{10!} .$

Recall that $\sqrt[a]{b} = b^{1/a}$ , then we can rewrite the above as

$\large (9!)^{1/9} \quad ? \quad (10!)^{1/10}$

To remove the fractions, let's raise both sides by the power $9 \times 10$ ,

$\large (9!)^{9\times10/9} \quad ? \quad (10!)^{9\times10/10}$

which simplifies to

$\large (9!)^{10} \quad ? \quad (10!)^{9}$

and is equivalent to

$\large (9!)^{9} \times (9!) \quad ? \quad (9! \times10)^{9} .$

or $\large (9!)^{9} \times 9! \quad ? \quad (9!) ^{9} \times 10^9 .$

Removing both sides of by $(9!)^9$ gives

$\large \bcancel{(9!)^{9}} \times 9! \quad ? \quad \bcancel{(9!)^{9}} \times 10^9 .$

This leaves us with

$1\times2\times\cdots\times9 \quad ? \quad 10\times10\times10\times \cdots\times10 .$

Since there are an equal number of integers being multiplied (in both sides), and that all the integers on the right hand are larger than those of the left hand side, then $LHS < RHS$ .

Hence, $9! \quad <\quad 10^9$ , or equivalently, $\large \sqrt[9]{9!} \quad < \quad \sqrt[10]{10!}$ .

Consider $\dfrac {\sqrt[10]{10!}}{\sqrt[9]{9!}} = (9!)^{\frac 1{10}-\frac 19} 10^{\frac 1{10}} = (9!)^{-\frac 1{90}} 10^{\frac 1{10}} = \dfrac {10^{\frac 1{10}}}{(9!)^\frac 1{90}} > \dfrac {10^{\frac 1{10}}}{(10^9)^\frac 1{90}} = \dfrac {10^\frac 1{10}} {10^\frac 1{10}} = 1$ . Therefore, $\dfrac {\sqrt[10]{10!}}{\sqrt[9]{9!}} > 1 \implies \boxed{\sqrt[10]{10!}} > \sqrt[9]{9!}$ .

(9!)^(1/9) ...... (10!)^(1/10), Rise to the power (90) for both sides, then (9!)^10 ...........(10!)^9, we have (9!)^9 x (9!) .....(9!)^9 x (10)^9, divided both by (9!)^9, (9!)< (10)^9, so (9!)^(1/9) < (10!)^(1/10)

(9!)^(1/9) =4.1...... (a)

and (10!)^(1/10)=4.5.................(b)

so,b>a

By AM - GM

$\sqrt[9]{9!} < \cfrac{1 + 2 + 3 + \cdots + 9}{9} \\ \sqrt[10]{10!} < \cfrac{1 + 2 + 3 + \cdots + 10}{10} \\ \sqrt[9]{9!} \times \sqrt[10]{10!} < \cfrac{1 + 2 + 3 + \cdots + 9}{9} \times \cfrac{1 + 2 + 3 + \cdots + 10}{10} = 2475 < \sqrt[10]{(10!)^2 } < \cfrac{1 + 2 + 3 + \cdots + 10}{10} \times \cfrac{1 + 2 + 3 + \cdots + 10}{10} \\ \implies \sqrt[9]{9!} < \sqrt[10]{(10!)}$

Let $f(x)=\Gamma(x+1)^\frac{1}{x}$

By calculus, we can show that $f'(x)>0$ for all positive x.

So $f(10)>f(9)$

First conisder $x=(9!)^{10}$ and $y=(9!)^{10}$

$x=(9!)^{10}=9!^9.9!$ and $y=(10!)^9=9!^9.10^9$

$9!=1 \times 2 \times \cdots \times 9$

$10^9=10 \times 10 \times \cdots \times 10$

Obviously , $9! <10^9$ .So $9!^9.9! < 9!^9.10^9 \implies x <y \implies (9!)^{10} <(10!)^{9} \implies \sqrt[9]{9!} <\sqrt[9]{10!}$

This is NOT a mathematically rigorous solution, but you can use a rough approximation of Stirling's formula...

n! ~ n^n --> (n!)^(1/n) ~ n

Since this increases with n, we'd expect by this heuristic that (10!)^1/10 > (9!)^(1/9)

The lack of rigor comes from (1) not using the full Stirling's approximation (I ignored the exponential factor, the radical factor, and the constants), and more importantly (2) Stirling's only holds in the limit as n --> infty, so there's no guarantee that it will even be close to accurate for small n. But this method is a 'quick and dirty' way to get the correct answer to this problem.