2003 Vietnamese Mathematical Olympiad

1/ • For $\alpha =-1$, it is easily seen that $x_n=0\forall n=1,2,3,\dots$. • For $\alpha \ne -1$, we have $x_n\ne -\alpha \forall n=1,2,3,\dots$, so we can write the inductive relation defining the sequence $\{x_n\}$ in the equivalent form: $$x_{n+1}=\frac{\alpha +1}{x_n+\alpha }\forall n=1,2,3,\dots (\ast )$$ From $x_1=0$, () yields $x_2=\frac{\alpha +1}{\alpha }$. Put $p_1=0$, $p_2=\alpha +1$, $q_1=1$, $q_2=\alpha$, then $x_1=\frac{p_1}{q_1}$, $x_2=\frac{p_2}{q_2}$. Suppose that $x_k$ has the form $x_k=\frac{p_k}{q_k}$, then () shows that $x_{k+1}$ has the form $x_{k+1}=\frac{p_{k+1}}{q_{k+1}}$, where $p_{k+1}=(\alpha +1)q_k$ and $q_{k+1}=\alpha q_k+p_k$. Therefore, by induction, $x_n$ has the form $x_n=\frac{p_n}{q_n}\forall n=1,2,3,\dots$, where $\{p_n\}$ and $\{q_n\}$ are defined by: $$p_1=0,q_1=1,q_2=\alpha ,p_{n+1}=(\alpha +1)q_n,q_{n+1}=\alpha q_n+p_n\forall n=1,2,3,\dots$$ From these relations, it follows that the sequence $\{q_n\}$ is defined by: $$q_1=1,q_2=\alpha \text{and}{q}_{n+1}=\alpha q_n+(\alpha +1)q_{n-1}\forall n=2,3,\dots$$ The characteristic equation of the sequence $\{q_n\}$ is $$x^2-\alpha x-(\alpha +1)=0.$$ This equation has two roots: $x_1=-1$ and $x_2=\alpha +1$. Therefore: $$\text{- If }\alpha =-2\text{ then }{q}_n=(-1{)}^{n-1}+(-1{)}^{n-1}(n-1)\forall n=1,2,3,\dots$$ $$\text{- If }\alpha \ne -2\text{ then }{q}_n=\frac{(-1{)}^{n-1}+(\alpha +1{)}^n}{\alpha +2}\forall n=1,2,3,\dots$$ and so: $$\text{- If }\alpha =-2\text{ then }{p}_1=0,p_n=-[((-1{)}^{n-2}+(-1{)}^{n-2})(n-2)]\forall n\ge 2.$$ - If $\alpha \ne -2$ then $p_1=0$, $p_n=\frac{(-1{)}^{n-2}+(\alpha +1{)}^{n-1}}{\alpha +2}(\alpha +1)\forall n\ge 2$. Consequently: - If $\alpha =-2$ then $x_n=\frac{n-1}{n}\forall n=1,2,3,\dots$. - If $\alpha \ne -2$ then $x_n=\frac{|(-1{)}^{n-2}+(\alpha +1{)}^{n-1}|(\alpha +1)}{(-1{)}^{n-1}+(\alpha +1{)}^n}\forall n=1,2,3,\dots$.

2/ • It is easy to see that: - If $\alpha =-1$ then $\lim x_n=0$. - If $\alpha =-2$ then $\lim x_n=1$. • When $\alpha \ne -2$, we have: $$x_{2k-1}=\frac{(\alpha +1{)}^{2k-1}-(\alpha +1)}{1+(\alpha +1{)}^{2k-1}}\forall k\ne 1,2,3,\dots$$ $$x_{2k}=\frac{(\alpha +1)+(\alpha +1{)}^{2k}}{(\alpha +1{)}^{2k}-1}\forall k=1,2,3,\dots$$ It follows that: - If $|\alpha +1|>1$ then $\lim x_{2k-1}=1$ and $\lim x_{2k}=1$. - If $|\alpha +1|<1$ then $\lim x_{2k-1}=-(\alpha +1)$ and $\lim x_{2k}=-(\alpha +1)$. In consequence, the sequence $\{x_n\}$ is convergent and: - If $|\alpha +1|>1$ then $\lim x_n=1$. - If $|\alpha +1|<1$ then $\lim x_n=-(\alpha +1)$.