A quite elegant and classic exercise of Calculus (in infinite series) is the following:
If the non-negative sequence $\{a_n\}$ is decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $na_n\to 0$.
To show this observe that, if $\,\sum_{n\ge n_0}a_n<\varepsilon/2$, then for for $n\ge 2n_0+1$, $$\frac{\varepsilon}{2}>a_{\lfloor n/2\rfloor}+\cdots+a_n\ge \frac{1}{2}na_n\ge 0.$$
This, in a sense, is related to the fact that $\sum\frac{1}{n}=\infty$. To go one step further, since $\sum\frac{1}{n\log n}=\infty$, can we obtain, with the same assumptions on $\{a_n\}$, that $\,n\log n\, a_n\to 0$?
This conjecture holds with the additional assumption that $b_n=na_n$ is also decreasing. To see this, let $n_0\in\mathbb N$, such that $\sum_{n\ge n_0}a_n<\varepsilon$. Then for $n\ge n_0^2$, we have $$ \varepsilon>\sum_{\sqrt{n}\le k\le n}a_n\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor}\sum_{k=\lfloor2^{\ell-1}\log_2\sqrt{n}\rfloor+1}^{ \lfloor2^\ell\log_2\sqrt{n}\rfloor}a_k\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor} (2^{\ell-1}\log_2\sqrt{n}-1)a_{\lfloor2^\ell\log_2\sqrt{n}\rfloor} \\ \ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor} \Big(\frac{1}{2}{\lfloor2^\ell\log_2\sqrt{n}\rfloor}-1\Big) a_{{\lfloor2^\ell\log_2\sqrt{n}\rfloor}}\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor}\Big(\frac{1}{2}n-1\Big)a_n\ge (\log_2 n-1)\Big(\frac{1}{2}n-1\Big)a_n. $$ Hence, $(\log_2 n-1)\big(\frac{1}{2}n-1\big)a_n\to 0$, which implies that $\,\,n\log n\,a_n\to 0$.
One step further: If $a_n>0$, $\sum_{n=1^\infty}a_n<0$ $a_n$, and the sequences $na_n$ and $n\log n a_n$ are decreasing, then $$ n\log n \log\log n\,a_n\to 0. $$