If $\sup_n \frac{n}{|b_n|}<\infty$ then $\limsup_{t\to\infty} \frac{\# \{n\ge 1:|b_n|\le t\}}{t}<\infty$

Question

Let $\{b_n\}$ be a non-zero sequence of complex numbers, $n\in \mathbb{N}^*$. According to a artical, if $\sup_n \frac{n}{|b_n|}<\infty$ then $\limsup_{t\to\infty} \frac{\# \{n\ge 1:|b_n|\le t\}}{t}<\infty$. Can you explain to me this?

Answer 1

Introduce $M=\sup\limits_{n}\frac{n}{|b_{n}|}$, then $M$ is finite and $|b_{n}|\geq \frac{n}{M}$ for all $n$. Thus, $$\frac{\#\{n\geq 1:|b_{n}|\leq t\}}{t}\leq \frac{\#\{n\geq 1:n\leq Mt\}}{t}\leq M$$ for every $t>0$, so $$\sup\limits_{t>0}\frac{\#\{n\geq 1:|b_{n}|\leq t\}}{t}\leq M$$ is finite, in particular, $$\limsup\limits_{t\to\infty}\frac{\#\{n\geq 1:|b_{n}|\leq t\}}{t}\leq\sup\limits_{t>0}\frac{\#\{n\geq 1:|b_{n}|\leq t\}}{t}$$ is finite. To sum up, $$\limsup\limits_{t\to\infty}\frac{\#\{n\geq 1:|b_{n}|\leq t\}}{t}\leq\sup\limits_{n}\frac{n}{|b_{n}|}.$$