Problem 1457

1. Understanding the Problem:
We are given a sequence of integers $u_k$ and a function $V_n$ which counts the number of terms in the sequence $u_k$ that are less than or equal to $n$. We need to show that if the series $\sum_{k=1}^{\infty} \frac{1}{u_k}$ converges, then the limit of $\frac{V_n}{n}$ as $n$ approaches infinity is zero.

2. Assumption and Goal:
We assume that $\sum_{k=1}^{\infty} \frac{1}{u_k} < \infty$. Our goal is to show that $\lim_{n \to \infty} \frac{V_n}{n} = 0$.

3. Using the Convergence of the Series:
Since $\sum_{k=1}^{\infty} \frac{1}{u_k}$ converges, for any $\varepsilon > 0$, there exists an integer $N$ such that for all $n \geq N$, the sum of the reciprocals of $u_k$ with $n < u_k \leq 2n$ is less than $\varepsilon$. This is because the tail of a convergent series can be made arbitrarily small.

4. Bounding the Number of Terms:
Consider the terms $u_k$ such that $n < u_k \leq 2n$. Each of these terms contributes at least $\frac{1}{2n}$ to the sum. Therefore, if there are $V_{2n} - V_n$ such terms, we have:
$$(V_{2n} - V_n) \cdot \frac{1}{2n} \leq \varepsilon$$
Simplifying this inequality, we get:
$$V_{2n} - V_n \leq 2n \varepsilon$$

5. Summing Dyadically:
We can now sum this inequality over dyadic intervals. For sufficiently large $n$, we can write:
$$V_{2n} \leq V_n + 2n \varepsilon$$
Repeating this process, we get:
$$V_{4n} \leq V_{2n} + 4n \varepsilon \leq V_n + 2n \varepsilon + 4n \varepsilon = V_n + 6n \varepsilon$$
Continuing this process, we see that for any $k$:
$$V_{2^k n} \leq V_n + (2n + 4n + 8n + \cdots + 2^k n) \varepsilon$$
The sum inside the parentheses is a geometric series with sum $(2^{k+1} - 2)n$. Therefore:
$$V_{2^k n} \leq V_n + (2^{k+1} - 2)n \varepsilon$$

6. Taking the Limit:
Dividing both sides by $2^k n$, we get:
$$\frac{V_{2^k n}}{2^k n} \leq \frac{V_n}{2^k n} + \frac{(2^{k+1} - 2)n \varepsilon}{2^k n} = \frac{V_n}{2^k n} + 2 \varepsilon - \frac{2 \varepsilon}{2^k}$$
As $k \to \infty$, the term $\frac{V_n}{2^k n}$ approaches zero, and we are left with:
$$\lim_{k \to \infty} \frac{V_{2^k n}}{2^k n} \leq 2 \varepsilon$$
Since $\varepsilon > 0$ was arbitrary, we conclude that:
$$\lim_{n \to \infty} \frac{V_n}{n} = 0$$
This completes the proof.

$\blacksquare$