Series in Real Analysis
Outline: One direction is easy comparison.
For the other, suppose that $\sum \frac{a_n}{1+a_n}$ converges. Then after a while $a_n\le 1$. For such $n$, we have $a_n\le \frac{2a_n}{1+a_n}$.
if $\sum a_n$ converge then $a_n\to 0$, so $a_n\sim \dfrac{a_n}{1+a_n}$, hence $\sum\dfrac{a_n}{1+a_n}$ converge.
if $\sum\dfrac{a_n}{1+a_n}$ converge, put $v_n=\dfrac{a_n}{1+a_n}$, then $a_n=\dfrac{v_n}{1-v_n}$, as $v_n\to 0$ we have $u_n\sim v_n$ hence $\sum a_n$ converge.