And suppose that $a_n > 0\quad \forall n$. Since the case $b_n := \frac{a_n}{1+a_n} \to p$ with $p \ne 0$ is clear, I'm supposing that $b_n\to 0$ and that $\sum b_n$ converges.
So for any real $M > 0$ I have that there's a natural $N$ such that $n\ge N \implies$ $$1+\frac{1}{a_n}>M$$
I feel that there's a way to limit $a_n$ by $b_n$ so that convergence of $\sum b_n$ implies the contradiction that $\sum a_n$ converges. But I'm getting nothing from these hypothesis.
If $a_n\not\to0$, then ${a_n\over1+a_n}\not\to0$. If $a_n\to0$ (and $a_n\ge0$), then ${a_n\over1+a_n}\gt{a_n\over2}$ for all large $n$. In either case $\sum{a_n\over1+a_n}$ diverges.
There are $2$ cases to consider:
$a)$ The sequence $\{a_n\}$ is bounded, which means $|a_n| < M$ for some $M > 0$. Thus $\dfrac{a_n}{1+a_n}> \dfrac{a_n}{1+M}$, and $\sum_{n}\dfrac{1} {1+M}a_n$ diverges to $+\infty$, hence $\sum_{n}\dfrac{a_n}{1+a_n}$ diverges to $+\infty$
$b)$ The sequence $\{a_n\}$ is unbounded, thus for each $k \ge 1$, we can find a term $a_{n_k}$ such that $a_{n_k}> k\implies \sum_{n}\dfrac{a_n}{1+a_n} \ge \sum_{k}\dfrac{a_{n_k}}{1+a_{n_k}} > \sum_{k}\dfrac{k}{1+k} > \sum_{k}\dfrac{1}{k} = +\infty$
In either case, we find that the series in question diverges by comparison with a diverging series.
asymptotic comparison
If $\lim_{n\to+\infty}a_n=0$ then
$$a_n\sim \frac {a_n}{1+a_n} \;\;(n\to+\infty) $$
so the two series have the same nature.
