# If $(a_n),\ (b_n)$ are positive decreasing sequences, $(a_n)$ is convex, $\sum a_n$ converges and $\sum b_n$ diverges, then $\frac{a_n}{b_n}\to 0.$

Start by considering this question. Is an attempt to generalise the role of $$\ f_1(n)=n\$$ to $$\ f_2(n)=n\log n\$$ and beyond, in the limit $$\ \displaystyle\lim_{n\to\infty} n a_n \to 0\$$ - so now we are considering whether $$\ \displaystyle\lim_{n\to\infty} (n\log n) a_n \to 0,\$$ I came up with the following proposition:

Proposition $$\ 1:\$$ If $$\ (a_n)_n,\ (b_n)_n,\$$ are positive decreasing sequences such that $$\ \displaystyle\sum a_n \$$ converges and $$\ \displaystyle\sum b_n \$$ diverges, then $$\ \frac{a_n}{b_n}\to 0.\$$

The following is a counter-example:

$$b_n = \frac{1}{n\log n},$$

$$\text{ For each } k\in\mathbb{N},\ \text{ let } a_n = \frac{ 1 }{ \left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right) } \text{ for all }\ 2^{2^k} < n\leq 2^{2^{k+1}} = \left(2^{2^{k}}\right)^2 .$$



In fact, Proposition 1 with $$b_n = \frac{1}{n\log n},$$ is the same as this question, which has similar counter-examples.

Proposition $$\ 2:\$$ If $$\ (a_n)_n,\ (b_n)_n,\$$ are positive decreasing sequences, $$\ (a_n)_n\$$ is convex, that is, $$\ a_n - a_{n+1} \geq a_{n+1} - a_{n+2}\quad \forall\ n,\$$ and $$\ \displaystyle\sum a_n \$$ converges and $$\ \displaystyle\sum b_n \$$ diverges, then $$\ \frac{a_n}{b_n}\to 0.\$$

Is this true or false? Tools that could be relevant:

Cauchy's Condensation test, in particular, Schlömilch's_Generalization.

The integral test

Stolz–Cesàro_theorem

The answer is no. But the type of counter-example compared to proposition $$1$$ is different in that this time we let $$a_n$$ be any function satisfying the properties, and then we choose $$b_n$$ based on $$a_n.$$ So for example, the following is a counter-example:
$$a_n = \frac{1}{n^2},$$
$$\text{ For each } k\in\mathbb{N},\ b_n = \frac{ 1 }{ \left({{2^2}^k}\right)^2 } \text{ if }\ {2^2}^k \leq n < {2^2}^{k+1} = \left({2^2}^k\right)^2.$$
The next natural question is: what if both $$\ (a_n)\$$ and $$\ (b_n)\$$ are required to be convex? I have asked this as another question.