IMC 2026 · Day 1 · P4

Let $y_{n} = \frac{x_{1} + x_{2} + \cdots + x_{n}}{n}$. Then we have the recurrence relation: $$\begin{equation} y_{n+1} = \frac{n y_{n} + x_{n+1}}{n+1}. \tag{1} \end{equation}$$ where $x_{n+1} = \arctan y_{n}$. By mathematical induction, $x_{n} > 0$ holds for all positive integers $n$.

Since $\arctan t < t$ for all $t > 0$, it follows that $x_{n+1} = \arctan y_{n} < y_{n}$. Substituting into equation (1): $$y_{n+1} < \frac{n y_{n} + y_{n}}{n+1} = \frac{(n+1) y_{n}}{n+1} = y_{n},$$ which implies $\left\{ y_{n} \right\}$ is strictly decreasing. As $y_{n} > 0$ for all $n$, by the Monotone Convergence Theorem, $\left\{ y_{n} \right\}$ converges. Let $\lim_{n \rightarrow \infty} y_{n} = y$.

Furthermore, since $y_{n} > y_{n+1}$, we have $$x_{n+1} = \arctan y_{n} > \arctan y_{n+1} = x_{n+2},$$ so $\left\{ x_{n} \right\}$ is also strictly decreasing and bounded below by 0. By the Monotone Convergence Theorem, $\left\{ x_{n} \right\}$ converges. Let $\lim_{n \rightarrow \infty} x_{n} = x$.

Taking limits on both sides of the recurrence: $$x = \lim_{n \rightarrow \infty} x_{n+1} = \lim_{n \rightarrow \infty} \arctan \left( \frac{x_{1} + x_{2} + \cdots + x_{n}}{n} \right) = \arctan y.$$

It is obvious that $y = x$. By the preservation of inequalities under limits, $x \geq 0$. If $x > 0$, then $$x = \arctan y = \arctan x < x,$$ which is a contradiction. Therefore $x = 0$, and hence $y = 0$.

Note that $$\frac{x_{n}}{y_{n}} = \frac{\arctan y_{n}}{y_{n}} \rightarrow 1 \text{ as } n \rightarrow \infty,$$ so $x_{n}$ and $y_{n}$ are equivalent infinitesimals as $n \rightarrow \infty$.

By Stolz's Theorem ($\infty / \infty$ form): $$\lim_{n \rightarrow \infty} \frac{1 / x_{n}^{2}}{\ln n} = \lim_{n \rightarrow \infty} \frac{1 / y_{n}^{2}}{\ln n} = \lim_{n \rightarrow \infty} \frac{\frac{1}{y_{n+1}^{2}} - \frac{1}{y_{n}^{2}}}{\ln (n+1) - \ln n}.$$

Simplify the numerator: $$\frac{1}{y_{n+1}^{2}} - \frac{1}{y_{n}^{2}} = \frac{y_{n}^{2} - y_{n+1}^{2}}{y_{n}^{2} y_{n+1}^{2}} = \frac{\left( y_{n} - y_{n+1} \right) \left( y_{n} + y_{n+1} \right)} {y_{n}^{2} y_{n+1}^{2}}.$$

As $n \rightarrow \infty$, $y_{n+1} \sim y_{n}$, so $y_{n} + y_{n+1} \sim 2 y_{n}$ and $y_{n}^{2} y_{n+1}^{2} \sim y_{n}^{4}$.

For the denominator: $$\ln (n+1) - \ln n = \ln \left( 1 + \frac{1}{n} \right) \sim \frac{1}{n} \quad (n \rightarrow \infty).$$

Thus $$\lim_{n \rightarrow \infty} \frac{\frac{1}{y_{n+1}^{2}} - \frac{1}{y_{n}^{2}}}{\ln (n+1) - \ln n} = \lim_{n \rightarrow \infty} \frac{\left( y_{n} - y_{n+1} \right) \cdot 2 y_{n}}{y_{n}^{4}} \cdot n = 2 \lim_{n \rightarrow \infty} \frac{n \left( y_{n} - y_{n+1} \right)}{y_{n}^{3}}.$$ Now compute $y_{n} - y_{n+1}$: $$y_{n} - y_{n+1} = y_{n} - \frac{n y_{n} + x_{n+1}}{n+1} = \frac{(n+1) y_{n} - n y_{n} - x_{n+1}}{n+1} = \frac{y_{n} - x_{n+1}}{n+1} = \frac{y_{n} - \arctan y_{n}}{n+1}.$$ Substitute back into the limit: $$2 \lim_{n \rightarrow \infty} \frac{n \cdot \frac{y_{n} - \arctan y_{n}}{n+1}}{y_{n}^{3}} = 2 \lim_{n \rightarrow \infty} \frac{n}{n+1} \cdot \frac{y_{n} - \arctan y_{n}}{y_{n}^{3}}.$$ Since $\frac{n}{n+1} \rightarrow 1$ as $n \rightarrow \infty$, and using the Taylor expansion $t - \arctan t \sim \frac{t^{3}}{3}$ for $t \rightarrow 0^{+}$: $$2 \lim_{t \rightarrow 0^{+}} \frac{t - \arctan t}{t^{3}} = 2 \cdot \frac{1}{3} = \frac{2}{3}.$$ Therefore $$\lim_{n \rightarrow \infty} \frac{1 / x_{n}^{2}}{\ln n} = \frac{2}{3}$$ and we conclude $$\lim_{n \rightarrow \infty} x_{n} \sqrt{\ln n} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$$