$\sum_{i=1}^{\infty}\frac{1}{a_i}$ converges.,prove that $\lim_{n \to \infty}\frac{b_n}{n}=0$.

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?

• You have tagged this "contest-math". Which contest, please? – Gerry Myerson Nov 23 '18 at 6:30
• Putnam contest 1964 B1 – mathpadawan Nov 23 '18 at 6:31
• I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky. – James Nov 23 '18 at 6:32

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 $$n \to \infty$$.