Prove that if $a_n\ge 0$ and $\sum_{n=1}^{\infty} a_n$ diverges. Then $\sum_{n=1}^{\infty} \dfrac{a_n}{1+a_n}$ also diverges.
I know that if $\sum_{n=1}^{\infty} a_n$ diverges then we do not know if $\sum_{n=1}^{\infty} \dfrac{a_n}{1+na_n}$ converges or diverges but what can we say about $\sum_{n=1}^{\infty} \dfrac{a_n}{1+a_n}$?
Hint: Either $(a_n)$ is bounded or it isn't. You may want to treat these cases separately.
Set $$ S=\{k\in\mathbb N: a_k>1 \}. $$
Case 1. If $S$ is infinite, then for every $k\in S$ $$ \frac{a_k}{1+a_k}=1-\frac{1}{1+a_k}>1-\frac12=\frac12, $$ and hence $$ \sum_{n=1}^\infty\frac{a_n}{1+a_n}\ge \sum_{k\in S}\frac{a_k}{1+a_k}\ge \sum_{k\in S}\frac12 =\infty. $$ Case 2. If $S$ is finite and $N=\max S$, then for every $k\in S$ $$ \frac{a_k}{1+a_k}\ge\frac{a_k}{2} $$ and hence we have $$ \sum_{n=1}^\infty \frac{a_n}{1+a_n}\ge \sum_{n=N+1}^\infty \frac{a_n}{1+a_n}\ge \sum_{n=N+1}^\infty \frac{a_n}{2}=\infty. $$
If $a_n$ does not tend to zero then neither does $a_n/(1+a_n)$. If otoh $a_n\to0$ then $a_n<1$ except for finitely many values of $n$, hence $a_n/(1+a_n)\ge a_n/2$ except for finitely many values of $n$.
