A seemingly true claim?

$\lim_{x \to \infty} \frac x{1+x} =1$. So if $a_{n_k} \to \infty$ then $\frac {a_{n_k}} {1+a_{n_k}} \to 1$.

Since $\frac {a_n} {1+a_n} <1$ for all $n$ we can conclude that the limit superior is $1$.

Just sharing my opinion.Assume the contrary that, $x_n=\frac{a_n}{1+a_n}$ tends to zero, then for any fixed $0<\epsilon<1$ there is a $N\in\Bbb N$ such that, $x_n<\epsilon,\forall n\ge N$, which gives $\displaystyle a_n<\frac{\epsilon}{1-\epsilon},\forall n\ge N$. Now take $M=max\{\frac{\epsilon}{1-\epsilon},a_1,...,a_{N-1}\}$, we see $a_n<M,\forall n\in\Bbb N$ i.e. Bounded-a contradiction that $\{a_n\}$ is not bounded. So the sequence $\{x_n\}$ doesnot tends to zero.