Suppose that $a_n > 0$ for all $n \in \mathbb{N}$ and that $\sum_{n=1}^\infty a_n = +\infty$. Let $b_n \colon= {a_n \over {1+na_n}}$ for all $n \in \mathbb{N}$.
Then we can show the following fact:
If the sequence $\{na_n\}_{n\in \mathbb{N}}$ is bounded above or if this sequence has a positive lower bound (of course this sequence is bounded below by $0$), then $\sum_{n=1}^\infty b_n = +\infty$.
However, can we come up with a case where neither of the above two hypotheses about $\{na_n\}_{n\in \mathbb{N}}$ holds (although $\sum_{n=1}^\infty a_n = +\infty$ still holds), but nevertheless $\sum_{n=1}^\infty b_n = +\infty$?
The idea is to define $a_n$ in such a way that on odd integers, $na_n$ is bounded above but not below by a positive number, and on even integers it is not bounded above, but it is bounded below by some positive number. For exemple, take $$a_n = \begin{cases}n &\text{if }n \text{ is odd,} \\ 1/n^2 &\text{if } n \text{ is even.}\end{cases}$$ It is obvious that $\sum_{n\ge 1} a_n = \infty$. Also, we have that $\liminf na_n = 0$ and $\limsup na_n = \infty$.
Now if we compute $$ \sum_{\substack{n\ge1 \\ n \text{ odd}}} b_n = \sum_{k\ge 0}\frac{2k+1}{4k^2 + 4k + 2} = \infty $$ This last claim is seen from the fact by using Cauchy's condensation criterion and the ratio test. Remark that adding the positive even terms will change nothing, hence $\sum_{n\ge 1} b_n = \infty$.