Romanian Mathematical Olympiad

We have $a_n > 0$ and $a_{n+1} < a_n$, for any $n \ge 1$. It turns out that the sequence $(a_n)_{n \ge 1}$ is convergent, with the limit $\ell \in [0, 1)$. From the recurrence relation, we obtain $\ell = \frac{\ell}{1+\sqrt{1+\ell}}$, so $\ell = 0$. Then $\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} (1 + \sqrt{1+a_n}) = 2$.

The recurrence relation implies $1 + a_{n+1} = \sqrt{1+a_n}$, for any $n \in \mathbb{N}^*$. Therefore, we have $\log_2(1+a_{n+1}) = \frac{1}{2}\log_2(1+a_n)$, for any $n \ge 1$. From $\log_2(1+a_1) = 1$ we obtain $\log_2(1+a_n) = \frac{1}{2^{n-1}}$, for all positive integers $n$ (geometric progression with the first term 1 and the ratio $1/2$). Hence $\lim_{n \to \infty} \sum_{k=1}^{n} \log_2(1+a_k) = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{2^{k-1}} = 2$.