Saudi Arabia Mathematical Competitions

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 $1,2,\dots ,n$. We can write $$S_{n-1}=n!\left(1+\frac{1}{2}+\dots +\frac{1}{n}\right)=n!\cdot H_n$$ and the above inequality is equivalent to $x_n^2<\frac{1}{n!H_n}$, hence $x_n<\frac{1}{\sqrt{n!H_n}}$.