Solving an inequality involving factorials and an exponent: $\frac{(n!)^2}{(2n)!} \leqslant \frac{1}{500}$

Question

I wish to solve the following inequality.

$\frac{(n!)^2}{(2n)!}$ $\leqslant$ $\frac{1}{500}$

$\forall$ n $\in$ $N_+$

Any help would be appreciated!

Edit: I am not allowed a calculator for this question.

Answer 1

Taking cue from what Hanul Jeon has said, $2n \choose n$ is an increasing sequence and it is easy to prove it.

Now the best way I prefer to prove the inequality is simple trial and error.
Put n=5, you get $2n \choose n$$= 252$
Put n=6, you get $2n \choose n$$= 924$
So the solution is $n\ge 6$ and $n \in \mathbb{N^+}$.

P.S. I don't think you need a calculator for such minor calculations.

Answer 2

Following up on the other hints (well, answers), here's a proof that you do not need a calculator.

$$\begin{array}{ccccccccccccc}\require{enclose}&&&&&&\enclose{circle}1\\&&&&&1&&1\\&&&&1&&\enclose{circle}2&&1\\&&&1&&3&&3&&1\\&&1&&4&&\enclose{circle}6&&4&&1\\&1&&5&&10&&10&&5&&1\\1&&6&&15&&\enclose{circle}{20}&&15&&6&&1\\&7&&21&&35&&35&&21&&7\\&&28&&56&&\enclose{circle}{70}&&56&&28\\&&&84&&126&&126&&84\\&&&&210&&\enclose{circle}{252}&&210\\&&&&&462&&462\\&&&&&&\enclose{circle}{924}\\\end{array}$$