International Mathematics Competition
for University Students
2026

IMC2026: Day 1, Problem 4

Problem 4. Let \(\displaystyle x_{1} > 0\). Define the sequence \(\displaystyle \{x_{n}\}\) by the recurrence

\(\displaystyle x_{n+1}=\arctan\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \text{ for all }n\geq 1. \)

Find \(\displaystyle \lim\limits_{n \to \infty} x_{n} \sqrt{\ln n}\), where \(\displaystyle \ln x\) denotes the natural logarithm of \(\displaystyle x\).

Wanlong Han, Henan, China

Solution. Let \(\displaystyle y_{n}=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\). Then we have the recurrence relation:

\(\displaystyle y_{n+1}=\frac{n y_{n}+x_{n+1}}{n+1}. \tag{1}\)

where \(\displaystyle x_{n+1}=\arctan y_{n}\). By mathematical induction, \(\displaystyle x_{n}>0\) holds for all positive integers \(\displaystyle n\).

Since \(\displaystyle \arctan t<t\) for all \(\displaystyle t>0\), it follows that \(\displaystyle x_{n+1}=\arctan y_{n}<y_{n}\). Substituting into equation (1):

\(\displaystyle y_{n+1}<\frac{n y_{n}+y_{n}}{n+1}=\frac{(n+1) y_{n}}{n+1}=y_{n},\)

which implies \(\displaystyle \{y_{n}\}\) is strictly decreasing. As \(\displaystyle y_{n}>0\) for all \(\displaystyle n\), by the Monotone Convergence Theorem, \(\displaystyle \{y_{n}\}\) converges. Let \(\displaystyle \lim _{n \to \infty} y_{n}=y\).

Furthermore, since \(\displaystyle y_{n}>y_{n+1}\), we have

\(\displaystyle x_{n+1}=\arctan y_{n}>\arctan y_{n+1}=x_{n+2},\)

so \(\displaystyle \{x_{n}\}\) is also strictly decreasing and bounded below by 0. By the Monotone Convergence Theorem, \(\displaystyle \{x_{n}\}\) converges. Let \(\displaystyle \lim _{n \to \infty} x_{n}=x\).

Taking limits on both sides of the recurrence:

\(\displaystyle x=\lim_{n\to \infty }x_{n+1}=\lim_{n\to \infty }\arctan\left( {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\right) =\arctan y.\)

It is obvious that \(\displaystyle y=x\). By the preservation of inequalities under limits, \(\displaystyle x \geq 0\). If \(\displaystyle x>0\), then

\(\displaystyle x=\arctan y=\arctan x<x,\)

which is a contradiction. Therefore \(\displaystyle x=0\), and hence \(\displaystyle y=0\).

Note that

\(\displaystyle \frac{x_{n}}{y_{n}}=\frac{\arctan y_{n}}{y_{n}} \to 1 \text{ as } n \to \infty,\)

so \(\displaystyle x_{n}\) and \(\displaystyle y_{n}\) are equivalent infinitesimals as \(\displaystyle n \to \infty\).

By Stolz’s Theorem (\(\displaystyle \infty/\infty\) form):

\(\displaystyle \lim _{n \to \infty} \frac{1 / x_{n}^{2}}{\ln n}=\lim _{n \to \infty} \frac{1 / y_{n}^{2}}{\ln n}=\lim _{n \to \infty} \frac{\frac{1}{y_{n+1}^{2}}-\frac{1}{y_{n}^{2}}}{\ln (n+1)-\ln n} .\)

Simplify the numerator:

\(\displaystyle \frac{1}{y_{n+1}^{2}}-\frac{1}{y_{n}^{2}}=\frac{y_{n}^{2}-y_{n+1}^{2}}{y_{n}^{2} y_{n+1}^{2}}=\frac{\left(y_{n}-y_{n+1}\right)\left(y_{n}+y_{n+1}\right)}{y_{n}^{2} y_{n+1}^{2}} .\)

As \(\displaystyle n \to \infty\), \(\displaystyle y_{n+1} \sim y_{n}\), so \(\displaystyle y_{n}+y_{n+1} \sim 2 y_{n}\) and \(\displaystyle y_{n}^{2} y_{n+1}^{2} \sim y_{n}^{4}\).

For the denominator:

\(\displaystyle \ln (n+1)-\ln n=\ln \left( 1+{\frac {1}{n}}\right) \sim {\frac {1}{n}} \quad (n\to \infty ).\)

Thus

\(\displaystyle \lim _{n \to \infty} \frac{\frac{1}{y_{n+1}^{2}}-\frac{1}{y_{n}^{2}}}{\ln (n+1)-\ln n}=\lim _{n \to \infty} \frac{\left(y_{n}-y_{n+1}\right) \cdot 2 y_{n}}{y_{n}^{4}} \cdot n=2 \lim _{n \to \infty} \frac{n\left(y_{n}-y_{n+1}\right)}{y_{n}^{3}} .\)

Now compute \(\displaystyle y_{n}-y_{n+1}\):

\(\displaystyle y_{n}-y_{n+1}=y_{n}-{\frac {ny_{n}+x_{n+1}}{n+1}}={\frac {(n+1)y_{n}-ny_{n}-x_{n+1}}{n+1}}={\frac {y_{n}-x_{n+1}}{n+1}}={\frac {y_{n}-\arctan y_{n}}{n+1}}.\)

Substitute back into the limit:

\(\displaystyle 2 \lim _{n \to \infty} \frac{n \cdot \frac{y_{n}-\arctan y_{n}}{n+1}}{y_{n}^{3}}=2 \lim _{n \to \infty} \frac{n}{n+1} \cdot \frac{y_{n}-\arctan y_{n}}{y_{n}^{3}} .\)

Since \(\displaystyle \frac{n}{n+1} \to 1\) as \(\displaystyle n \to \infty\), and using the Taylor expansion \(\displaystyle t-\arctan t \sim \frac{t^{3}}{3}\) for \(\displaystyle t \to 0^{+}\):

\(\displaystyle 2 \lim _{t \to 0^{+}} \frac{t-\arctan t}{t^{3}}=2 \cdot \frac{1}{3}=\frac{2}{3} .\)

Therefore

\(\displaystyle \lim _{n \to \infty} \frac{1 / x_{n}^{2}}{\ln n}=\frac{2}{3}\)

and we conclude

\(\displaystyle \lim _{n \to \infty} x_{n} \sqrt{\ln n}=\sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}\)