IMC 2011 · Day 2 · P6

We prove by induction that $$0 < 1 - x_n < \frac{1}{2^{n+1}}.$$ Then we will have $(1 - x_n) \to 0$ and therefore $x_n \to 1$.

The case $n = 0$ is true since $\frac{1}{2} < x_0 = a_0 < 1$.

Supposing that the induction hypothesis holds for $n$, from the recurrence relation we get $$1 - x_{n+1} = 1 - \frac{a_{n+1} + x_n}{1 + a_{n+1} x_n} = \frac{1 - a_{n+1}}{1 + a_{n+1} x_n} (1 - x_n).$$ By $$0 < \frac{1 - a_{n+1}}{1 + a_{n+1} x_n} < \frac{1 - \frac{1}{2}}{1 + 0} = \frac{1}{2}$$ we obtain $$0 < 1 - x_{n+1} < \frac{1}{2} (1 - x_n) < \frac{1}{2} \cdot \frac{1}{2^{n+1}} = \frac{1}{2^{n+2}}.$$ Hence, the sequence converges in all cases and $x_n \to 1$.

As is well-known, $$\tanh(u + v) = \frac{\tanh u + \tanh v}{1 + \tanh u \tanh v}$$ for all real numbers $u$ and $v$.

Setting $u_n = \operatorname{ar\,tanh} a_n$ we have $x_n = \tanh(u_0 + u_1 + \dots + u_n)$. Then $u_0 + u_1 + \dots + u_n > (n+1) \operatorname{ar\,tanh} \frac{1}{2}$ and $\lim\limits_{n \to \infty} x_n = \lim\limits_{u \to \infty} \tanh u = 1$.

Remark. If the condition $a_n \in (\frac{1}{2}, 1)$ is replaced by $a_n \in (0, 1)$ then the sequence remains increasing and bounded, but the limit can be less than 1.