Question 55
For every positive integer $n$, let
$$
x_{n} = \frac{(2n+1) \cdot (2n+3) \cdots (4n-1) \cdot (4n+1)}{2n \cdot (2n+2) \cdots (4n-2) \cdot 4n}
$$
Prove that $\frac{1}{4n} \lt x_{n} - \sqrt{2} \lt \frac{2}{n}$.
$$
x_{n} = \frac{(2n+1) \cdot (2n+3) \cdots (4n-1) \cdot (4n+1)}{2n \cdot (2n+2) \cdots (4n-2) \cdot 4n}
$$
Prove that $\frac{1}{4n} \lt x_{n} - \sqrt{2} \lt \frac{2}{n}$.