Find $N\in\mathbb{N}$ such that ${(n!)^2 \over (2n - 1)!} \leq {1 \over 25000}$. Find $N$ belonging to $\mathbb{N}$ such that ${(n!)^2 \over (2n - 1)!} \leq {1 \over 25000}$.
I tried to find such an $N$, but didn't succeed; I tried making the left side a bigger but simpler expression, it didnt work out either.
Cf.

 A: The way I would go about it is to first note that
$$\frac{(n!)^2}{(2n-1)!} = \frac{n!}{(n+1)\times (n+2)\times \cdots \times(2n-1)} $$
And if you invert both sides, all you want is that
$$\frac{(n+1)\times (n+2)\times \cdots \times(2n-1)}{n!} \geq 25000$$
Also note that you don't need the tightest bound, you only need a bound that works. Can you get something from this last inequality?
All factors of the numerator, by themselves, are greater than any factors in the denominator. Try using that to your advantage.
A: You can just try small values of $n$. You'll find that $n=11$ works. Indeed, every $n\ge11$ works.
Or note that
$$
{(n!)^2 \over (2n - 1)!}
={{2n} \over \binom{2n}{n}}
$$
and use this elementary bound for the central binomial coefficient
$$
\frac{4^n}{2n+1} \leq {2n \choose n}
$$
We then need to find $n$ such that
$$
\frac{2n(2n+1)}{4^n} \le \frac{1}{25000}
$$
Since $2n(2n+1)$ is eventually less than $2^n$, it suffices to find $n$ such that
$$
\frac{1}{2^n} \le \frac{1}{25000}
$$
or $2^n \ge 25000$. This is true for all $n\ge 14$.
A: HINT.-$${(n!)^2 \over (2n - 1)!} \leq {1 \over 25000}\iff \frac{1}{2n}\cdot\frac{(2n)!}{n!\cdot n!}=\frac{1}{2n}\binom{2n}{n}\ge25000$$
so you can calculate 
You have $$\left\lfloor\frac{1}{20}\binom{20}{10}\right\rfloor=9237\lt25000\lt\left\lfloor\frac{1}{22}\binom{22}{11}\right\rfloor=32065$$Thus$$\color{red}{n\ge11}$$
