Consider the sequence of real numbers $(a_n)_{n \in \mathbb{N}}$ such that $a_n>0$ and such that the following series, $$ \sum_{n=1}^\infty \frac{a_n}{n} < \infty. $$ Is the sequence $(a_n)_{n \in \mathbb{N}}$ bounded?
The intuition says to me that it must be bounded because if we let it grow, then in someway it would be comparable to the harmonic series which we know it's divergent. But I can't find a nice proof to it or if there's some article or book to read about it.
Your intuition is not correct. Take $$a_n =\begin{cases}2^{k}&\text{if } n = 2^{2k}\\\frac1n & \text{otherwise}\end{cases}.$$
This means that $a_n$ becomes
$$1, \frac12, \frac13, 2, \frac15, \frac16, \frac17, \dots, \frac1{15},8,\frac1{17},\dots,\frac1{63},32,\frac1{65},\dots$$
Clearly, $a_n$ is not a bounded sequence. However, the sum $$\sum_{n=1}^\infty \frac{a_n}{n}$$ now becomes becomes an interweaved combination of two sums, one of them being $$\sum \frac1{n^2}$$ with a couple of terms missing, and the other one being $\sum 2^{-k}$.
Since both of them converge, so does their "combination".