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:
$$b_n = \frac{1}{n\log n},$$
$$\text{ For each } k\in\mathbb{N},\ \text{ let } a_n = \frac{ 1 }{ \left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right) } \text{ for all }\ 2^{2^k} < n\leq 2^{2^{k+1}} = \left(2^{2^{k}}\right)^2 .$$
$$$$
In fact, Proposition 1 with $b_n = \frac{1}{n\log n},$ is the same as this question, which has similar counter-examples.
Proposition $\ 2:\ $ If $\ (a_n)_n,\ (b_n)_n,\ $ are positive decreasing sequences, $\ (a_n)_n\ $ is convex, that is, $\ a_n – a_{n+1} \geq a_{n+1} – a_{n+2}\quad \forall\ n,\ $ and $\ \displaystyle\sum a_n \ $ converges and $\ \displaystyle\sum b_n \ $ diverges, then $\ \frac{a_n}{b_n}\to 0.\ $
Is this true or false? Tools that could be relevant:
Cauchy's Condensation test, in particular, Schlömilch's_Generalization.
Best Answer
The answer is no. But the type of counter-example compared to proposition $1$ is different in that this time we let $a_n$ be any function satisfying the properties, and then we choose $b_n$ based on $a_n.$ So for example, the following is a counter-example:
$$ 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.