bryan blog: Putnam 2001 B6 Problem

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Putnam 2001 B6 Problem

originally created on 2026-05-21

I have a friend, and this friend gives me fun problems. I will turn this into a series and post the interesting problems and solutions here.

For this problem, I know that this solution exists online. However, I want to try to solve it on my own first (with some internet assistance, LOL). I will also keep an update log for any corrections to the solution (as I may be wrong with my original blog post).

The Problem

The problem is as follows:

You have a sequence $a_1 < a_2 < a_3 < \ldots$ of positive real numbers such that $lim_{n \to \infty} \frac{a_n}{n} = 0$ .
Must there exist infinitely many positive integers $n$ such that

a_{n - i} + a_{n + i} < 2a_n \quad \forall i = 1, 2, \ldots, n - 1?

Intuition: Convexity

The first thing I saw was a midpoint inequality.

We can graph the points $(n, a_n)$ on the Cartesian plane. This creates a set $\{(n, a_n) | n \in \mathbb{Z}^+\}$ of points.
Meanwhile, we can look at the inequality: $a_{n - i} + a_{n + i} < 2a_n$ . This is equivalent to saying that the point $(n, a_n)$ is below the midpoint of the points $(n - i, a_{n - i})$ and $(n + i, a_{n + i})$ .
Using an example $y = \sqrt{x}$ , we can see that this applies. An image is provided below.

Graph of a sequence with the midpoint inequality

As we can see, the midpoint inequality is satisfied. The straight line is the midpoint of the two points, and the curve point is above the straight line. However, this is only an example, and we need to show that this is true for ALL sequences for infinitely many points.

To do this, we use the upper convex hull.

Upper Convex Hulls

To understand what a convex hull is, we first need to go over the convex set.

A set $S$ of points in a plane is convex if for any two points $A$ and $B$ in $S$ , the line segment that connects $A$ and $B$ is also in $S$ .

There are some other mathematical definitions, but let's keep it simple. Now, a convex hull of any set of points $S$ is the SMALLEST CONVEX SET that contains all of the points in $S$ .

Meanwhile, the upper convex hull of a set of points $S$ is the part of the convex hull that is above the points in $S$ .

Looking at the Upper Convex Hull

Going back to the problem, we have discrete points $(n, a_n)$ . We can look at the upper convex hull of these points and prove two things:

There are infinitely many points on the upper convex hull.
All points on the upper convex hull satisfy the midpoint inequality seen earlier.

To do the first part, we will use contradiction.

Assume there are only finitely many points on the upper convex hull. Let the largest point be $(N, a_N)$ .
The slope of the convex hull to any point $(n, a_n)$ where $n > N$ is $\sup_n\frac{a_n - a_N}{n - N}$ . This slope would hover above all subsequent points in the sequence.
We know that $lim_{n \to \infty} \frac{a_n}{n} = 0$ . This means that $lim_{n \to \infty} \frac{a_n - a_N}{n - N} = 0$ .
This means that the supremeum over these slopes is actually MAXIMIZED at some point $M$ that lies on the sequence, causing a contradiction.

There's our proof that there are infinitely many points on the upper convex hull. The second part is easier!

By the definition of the upper convex hull - a convex set - we know that any boundary point $(n, a_n)$ on the upper convex hull lies above the midpoint of any two points $(n - i, a_{n - i})$ and $(n + i, a_{n + i})$ since they are in the convex set.

Conclusion

Since both points are proven, we can conclude that there are infinitely many points that satisfy the problem's inequality. In other words, yes.

This was a fun problem! An update log will be provided below for any corrections to the solution. I will try to keep it as organized as possible!