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Algebra Difficulty 5.5 AIME, harder Prove it

For every positive integer nn, let

xn=(2n+1)⋅(2n+3)⋯⋯⋅(4n−1)⋅(4n+1)2n⋅(2n+2)⋯⋯⋅(4n−2)⋅4n x_{n}=\frac{(2 n+1) \cdot(2 n+3) \cdots \cdots \cdot(4 n-1) \cdot(4 n+1)}{2 n \cdot(2 n+2) \cdots \cdots \cdot(4 n-2) \cdot 4 n}

Prove that 14n<xn−2<2n\frac{1}{4 n}<x_{n}-\sqrt{2}<\frac{2}{n}.

## Solutions

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Solution

Squaring both sides of the given equality and applying x(x+2)⩽(x+1)2x(x+2) \leqslant (x+1)^{2} to the numerator of the obtained fraction and cancelling we have

xn2⩽(2n+1)⋅(4n+1)(2n)22+1n x_{n}^{2} \leqslant \frac{(2 n+1) \cdot (4 n+1)}{(2 n)^{2}}2 + \frac{1}{n}

Hence

1n2 and xn<2. The result then follows from the second chain of inequalities. \frac{1}{n}\sqrt{2} \text{ and } x_{n} < 2. \text{ The result then follows from the second chain of inequalities.}

Comment. These inequalities can easily be improved. For example, the inequalities in the solution involving xn2x_{n}^{2} can immediately be replaced by 32n<xn2−2<2n\frac{3}{2 n} < x_{n}^{2} - 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.