What is the fastest rate that $a_n$ can increase while still making $b_n \rightarrow 0$?

Question

Suppose we have $b_n = \frac{2 a_n}{2n + 3}$, where $a_n$ is a number depending on n. Then what is the fastest rate that $a_n$ can increase while still making $b_n \rightarrow 0$? My only thought is that we can apply L'Hospital's Rule such that $\lim_{n \rightarrow \infty} \frac{a'_n}{2} = 0$, i.e. $a_n$ increases less than linear. But I'm not sure what is the fastest rate it can increase.

Answer 1

This is the set of all sequences $a_n$, where for any $c>0$, you can find some $n_0$ such that for all $n\ge n_0$, $|a_n| <cn \text{ or } |a_n| < c(2n+3)$ here. This is a special definition of a set of functions/sequnces that is so much used (together with other notations) in analysis of algorithms' runtimes, that it is named $o(n)$ (Little o).

There are many sequences in this set like $\log n,\, \log^2 n, \,\sqrt{n}, \,n^\epsilon : \epsilon < 1, \,n^\epsilon\log^kn : \epsilon < 1, k\in\mathbb{N}$ etc., but $\sup(o(n))$ does not exist, because for any such fastest $a_n$ where $\frac{2 a_n}{2n + 3} \to 0$, we can find some $a'_n$ where $\frac{a_n}{a'_n} \to 0$ and $\frac{2 a'_n}{2n + 3} \to 0$

Answer 2

$a_n$ has to be $o(n)$ – if it is $\Theta(n)$ the limit will be some nonzero number. But $a_n=n^x$ is $o(n)$ for any $x<1$; we can have an infinite increasing sequence of such $x$, and those generate an infinite sequence of ever faster $a_n$ that all induce $\lim b_n=0$. There is thus no fastest growth rate that $a_n$ can have.