prove that $\sum_{n=1}^{\infty}\frac{a_{n}}{1+a_{n}}$ converges iff $\sum_{n=1}^{\infty}{a_{n}} $ converges.
I proved one direction: $\frac{a_{n}}{1+a_{n}}\leq a_n$ therefore if $\sum a_n$ converges, then by the comparison test so does the second sum..
Now, if $\sum \frac{a_{n}}{1+a_{n}}$ converges... how do I prove that $\sum a_n$ converges?
Supposing $a_n\ge 0$.
$$\sum\frac{a_{n}}{1+a_{n}}\text{ converges} \implies \lim\frac{a_{n}}{1+a_{n}}=0\implies \lim a_n =0\implies 1+a_n\le 2\text{ for $n$ large enough.}$$ Now, $$a_n\le\frac{2a_n}{1+a_n}\text{ for $n$ large enough.}$$ EDIT: the implication $\lim\frac{a_{n}}{1+a_{n}}=0\implies \lim a_n =0$ requires some work. If $\lim a_n =0$ is false, then some subsequence diverges to infinity or converges to some limit $\ne 0$.
If the original series converges, it means that $a_n \to_n 0$, hence $\frac{a_n}{1+a_n} \to_n 0$, because the numerator tends to 0 and denominator to 1. Now take the limit of the ratio $\frac{h(n)}{g(n)}$, where $h = a_n, \ g =\frac{a_n}{1+a_n}$. Clearly it tends to 1, therefore $g=O(h)$, they are of the same order, and therefore the second sum converges by comparison test.
EDIT: it fact, it's enough to just show the second argument to prove convergence of $g$.
