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Saudi Arabian Mathematical Competitions

Algebra Difficulty 5.8 AIME, harder Prove it Saudi Arabia

Let $n \geq 2$ be a positive integer and let $x_{n}$ be a positive real root to the equation $x(x+1) \ldots(x+n)=1$ . Prove that
$x_{n}<\frac{1}{\sqrt{n!H_{n}}}$
where $H_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n}$ .

Solution

Because $x_{n}$ satisfies the equation we have $x_{n}^{2} \cdot S_{n-1}< 1$ , where $S_{n-1}$ is the $n-1$ symmetric sum of $1,2, \ldots, n$ . We can write
$S_{n-1}=n!\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)=n!\cdot H_{n}$
and the above inequality is equivalent to $x_{n}^{2}<\frac{1}{n!H_{n}}$ , hence $x_{n}<\frac{1}{\sqrt{n!H_{n}}}$ .

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