IMO 2014 · Problem 1

Let \(a_0<a_1<a_2<\cdots\) be an infinite sequence of positive integers. Prove that there exists a unique integer \(n\geq 1\) such that \[ a_n<\frac{a_0+a_1+\cdots+a_n}{n}\leq a_{n+1}. \]

Official text of the problem. Official problem paper (PDF)

Solution

Write \(S_n=a_0+a_1+\cdots+a_n\), and for every \(n\geq1\) define \[ b_n=S_n-na_n. \] Then \(b_1=a_0>0\). Moreover, \[ b_{n+1} =S_n-na_{n+1} =b_n-n(a_{n+1}-a_n). \] Since the \(a_i\) are strictly increasing integers, \(a_{n+1}-a_n\geq1\). Consequently, \[ b_{n+1}\leq b_n-n<b_n, \] and repeated application gives \[ b_n\leq a_0-\sum_{k=1}^{n-1}k =a_0-\frac{n(n-1)}2. \] Thus \(b_n\leq0\) for all sufficiently large \(n\).

Let \(N\) be the least positive integer such that \(b_N\leq0\). Because \(b_1>0\), we have \(N\geq2\). Setting \(n=N-1\), we obtain \[ b_n>0,\qquad b_{n+1}\leq0. \] Using the definitions, these inequalities become \[ S_n-na_n>0,\qquad S_n-na_{n+1}\leq0. \] Dividing by \(n>0\) proves \[ a_n<\frac{S_n}{n}\leq a_{n+1}. \]

Finally, any index satisfying the required inequalities must satisfy \(b_n>0\) and \(b_{n+1}\leq0\). Since \((b_n)\) is strictly decreasing, this can occur only at \(n=N-1\). Hence the required integer exists and is unique. \(\square\)

The solution was checked against the official IMO shortlist solutions.