75th 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 $n\in {\mathbb{N}}^{\ast }$. 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 ${\lim }_{n\to \infty }{\sum }_{k=1}^n{\log }_2(1+a_k)={\lim }_{n\to \infty }{\sum }_{k=1}^n\frac{1}{2^{k-1}}=2$.