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Help: Let $a_1,a_2,...$ be positive integers such that $\sum_{i=1}^{\infty}\frac{1}{a_i}$ converges. For each $n$, let $b_n$ denote the number of positive integers $i$ for which $a_i \leq n$. prove that $\lim_{n \to \infty}\frac{b_n}{n}=0$.

Intuitively, $a_i$ should grow large fast enough for $1/a_i$ to converge. so gap between $a_i$ should be bigger and bigger? Is this the right idea?

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    $\begingroup$ You have tagged this "contest-math". Which contest, please? $\endgroup$ – Gerry Myerson Nov 23 '18 at 6:30
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    $\begingroup$ Putnam contest 1964 B1 $\endgroup$ – mathpadawan Nov 23 '18 at 6:31
  • $\begingroup$ I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky. $\endgroup$ – James Nov 23 '18 at 6:32
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Let $\epsilon >0$ and choose $N$ such that $\sum_{i=N}^{\infty} \frac 1 {a_i} <\epsilon$. Let $J_n=\{i>N:a_i \leq n\}$. Then $\sum_{i\in J_n} \frac 1 n\leq \sum_{i\in J_n} \frac 1 {a_i} <\epsilon$. Hence $\frac {card(J_n)} n <\epsilon$ for all $n$. Now $b_n \leq N+card(J_n)$ so $\frac {b_n} n \leq \epsilon +\frac N n \to \epsilon $ as $n \to \infty$.

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