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Algebra Difficulty 6.2 National olympiad Prove it

28. Given xn=(2n+1)(2n+3)⋯(4n−1)(4n+1)(2n)(2n+2)⋯(4n−2)(4n)x_{n}=\frac{(2 n+1)(2 n+3) \cdots(4 n-1)(4 n+1)}{(2 n)(2 n+2) \cdots(4 n-2)(4 n)}, prove: 14n<xn−\frac{1}{4 n}<x_{n}- 2<2n\sqrt{2}<\frac{2}{n}. (2001 Polish Mathematical Olympiad Problem)

Solution

28. Using the inequality (x+1)(−x−1)−(4n+1)2(2n)(4n)=16n2+8n+18n2>16n2+8n8n2=2+1n(x+1)(-x-1)-\frac{(4 n+1)^{2}}{(2 n)(4 n)}=\frac{16 n^{2}+8 n+1}{8 n^{2}}>\frac{16 n^{2}+8 n}{8 n^{2}}=2+\frac{1}{n}

From (2) we get xn2−2>0x_{n}^{2}-2>0, so,
xn−2=xn2−2xn+2xn2−22+2>1(2+2)n>14nx_{n}-\sqrt{2}=\frac{x_{n}^{2}-2}{x_{n}+\sqrt{2}}\frac{x_{n}^{2}-2}{2+\sqrt{2}}>\frac{1}{(2+\sqrt{2}) n}>\frac{1}{4 n}

In summary, 14n<xn−2<2n\frac{1}{4 n}<x_{n}-\sqrt{2}<\frac{2}{n}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.