THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

It is easily seen that the $x_n$ are all rational numbers greater than $1$. Rewrite the recurrence formula in the form $x_n=1/(x_{n+1}-1)-1/(x_n-1)$, $n\ge 1$, to get $$\sum _{k=1}^n{x}_k=\frac{1}{x_{n+1}-1}-\frac{1}{x_1-1}=\frac{x_n^2-x_n+1}{x_n-1}-3=\frac{(x_n-2{)}^2}{x_n-1}.$$ Finally, express the positive rational number $x_n-1$ in lowest terms, $x_n-1=a/b$, to deduce that $(a-b{)}^2/(ab)$ expresses ${\sum }_{k=1}^n{x}_k$ in lowest terms. The conclusion follows.