Positive series problem: $\sum\limits_{n\geq1}a_n=+\infty$ implies $\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty$

Proving the contrapositive statement seems cleaner to me. Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then ${a_n\over 1+{a_n}}\rightarrow 0$. This implies that ${a_n}$ is eventually less than one, so ${a_n\over2}\le {a_n\over a_n+1}$ for $n$ sufficiently large. The comparision test then shows that $\sum a_n$ converges.

I would argue by cases.

Case 1. $a_n\ge1$ for infinitely many $n\in\mathbb N$.

In this case, for each such $n$ we have $\frac{a_n}{1+a_n}=1-\frac{1}{1+a_n}\ge1-\frac12$, from which the claim easily follows.

Case 2. $a_n\ge1$ for only finitely many $n\in\mathbb N$.

In this case for every other $n$ we have $a_n<1$ and thus $\frac{a_n}{1+a_n}\ge\frac{a_n}{1+1}=\frac{a_n}2$. Since finitely many terms can't affect the convergence/divergence of a series, this will also diverge. (Since $\sum\limits_{n=1}^\infty\frac{a_n}{2}$ does.)

Alternatively, split the problem up in cases:

1, If there is a natural $N\in\mathbb{N}$ s.t $a_n\leq{1}$ for $n\geq N$, what can you conclude?

2, If (1) is not true, for every natural $N$ we can find a $n\geq{N}$ s.t $a_n>1.$ Now passing to a subsequence and comparing with a series with each term equal to a constant (more precisely $1/2$ or lower), the result follows.