Romanian Mathematical Olympiad

Obviously, $a_0=-1$. For each $n\in {\mathbb{N}}^{\ast }$, denote by $k_n$ the number of terms among $a_1,a_2,\dots ,a_n$ which are equal to $1$ (the other being $-1$). Then, the sequence $(x_n{)}_{n\ge 0}$ whose limit must be $\frac{1}{2}$ is $x_n=k_n\sqrt{n+1}-k_n\sqrt{n-1}$, or $$x_n=\frac{2k_n}{\sqrt{n+1}+\sqrt{n-1}}=\frac{2\frac{k_n}{\sqrt{n}}}{\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}}.$$ It is enough to construct $(a_n{)}_{n\ge 1}$ such that ${\lim }_{n\to \infty }\frac{k_n}{\sqrt{n}}=\frac{1}{2}$. (*) Let us choose $a_n=1$ if and only if $n=(2m{)}^2$, with $m\in {\mathbb{N}}^{\ast }$. Then $(2k_n{)}^2\le n<(2(k_n+1){)}^2$, whence $\frac{\sqrt{n}}{2}-1<k_n\le \frac{\sqrt{n}}{2}$, which guarantees (*).