Vijetnam, 2009

From its definition, it is easy to see that $x_n > 0$ for all $n \ge 1$.

Reformulate the defining relation for the sequence $(x_n)$ in the following form:
$$2x_n - x_{n-1} = \sqrt{x_{n-1}^2 + 4x_{n-1}} \quad \forall n \ge 2.$$
It follows that:
$$x_{n-1} = x_n^2 - x_n x_{n-1} \quad \forall n \ge 2.$$
Consequently:
$$\frac{1}{x_n^2} = \frac{1}{x_{n-1}} - \frac{1}{x_n} \quad \forall n \ge 2$$
(since $x_n \ne 0$ for all $n \ge 2$).

Hence, for all $n \ge 2$, we have
$$y_n = \sum_{i=1}^{n} \frac{1}{x_i^2} = \frac{1}{x_1^2} + \sum_{i=2}^{n} \left( \frac{1}{x_{i-1}} - \frac{1}{x_i} \right) = \frac{1}{x_1^2} + \frac{1}{x_1} - \frac{1}{x_n} = 6 - \frac{1}{x_n}$$
It follows easily that $\left(\frac{1}{x_n}\right)$ is a decreasing sequence, bounded below by $0$. Hence $\left(\frac{1}{x_n}\right)$ converges and according to the above, we have
$$\lim_{n \to \infty} \frac{1}{x_n} = 0.$$
Thus $(y_n)$ is convergent and
$$\lim_{n \to \infty} y_n = 6.$$