Problem 1841

Let $r$ denote the largest real number satisfying $\frac{b_n}{n^2} \geq r$ for all positive integers $n$, where $b_1, b_2, \dotsc$ are positive integers satisfying
$$1 = \frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb$$

We aim to determine the possible values of $r$.

### Claim 1: $r = \frac{1}{2}$ works and is maximal.
To achieve $r = \frac{1}{2}$, consider the sequence $b_n = \frac{n(n+1)}{2}$. Then,
$$\frac{b_n}{n^2} = \frac{n(n+1)}{2n^2} = \frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n},$$
which satisfies the condition $\frac{b_n}{n^2} \geq \frac{1}{2}$.

We can inductively show that $b_n \leq \frac{n(n+1)}{2}$. The base case is given. Assuming the hypothesis holds for all integers less than $n$, we have
$$\frac{b_n}{n^2} < \frac{b_{n-1}}{(n-1)^2} \leq \frac{n}{2(n-1)} \implies b_n < \frac{n^3}{2(n-1)}.$$
It is easy to verify that the largest possible $b_n$ is $\frac{n(n+1)}{2}$, as claimed.

### Claim 2: All $r < \frac{1}{2}$ work.
Consider the sequence $a_n := \left\lceil kn^2 \right\rceil + n$ for $k < \frac{1}{2}$. Since $a_n$ is $O(n^2)$, there exists $N$ such that for all $n \geq N$,
$$\frac{a_n}{n^2} < \frac{1}{2}.$$
Define the sequence $b_n$ as follows:
$$b_n := \begin{cases} \frac{n(n+1)}{2} & \text{for } n < N, \\ a_n & \text{for } n \geq N. \end{cases}$$
By definition of $N$, $\frac{b_n}{n^2} > \frac{b_{n+1}}{(n+1)^2}$ for $n < N$. For $n \geq N$, we want to show that
$$\frac{\left\lceil kn^2 \right\rceil + n}{n^2} > \frac{\left\lceil k(n+1)^2 \right\rceil + n + 1}{(n+1)^2}.$$
Since $\left\lceil kn^2 \right\rceil \geq kn^2$,
$$\frac{\left\lceil kn^2 \right\rceil + n}{n^2} \geq k + \frac{1}{n},$$
and since $\left\lceil k(n+1)^2 \right\rceil < k(n+1)^2 + 1$,
$$\frac{\left\lceil k(n+1)^2 \right\rceil + n + 1}{(n+1)^2} < k + \frac{n+2}{(n+1)^2}.$$
It suffices to verify that
$$\frac{1}{n} \geq \frac{n+2}{(n+1)^2} \iff (n+1)^2 \geq n(n+2),$$
which is true.

Combining these two claims, we conclude that the possible values of $r$ are:
$$0 \leq r \leq \frac{1}{2}.$$

The answer is: \boxed{0 \leq r \leq \frac{1}{2}}.