Divergence of $\sum \frac{a_j}{1+a_j}$ Given: $a_j >0$ and $\sum a_j$ diverges. Show that $\sum \frac{a_j}{1+a_j}$ diverges. 
Hint: show that if it converged, $a_j$ -> 0.
I don't understand how to think about this problem. Is there a convergence test I should use? I tried starting with the hint, but don't know how to conclude anything about the individual parts from the fraction's supposed convergence.
 A: Hint: To show the series diverges, notice that we have inequality $$\frac{a_{j}}{1+a_{j}}\geq\min\left(\frac{1}{2},\frac{a_{j}}{2}\right). $$ This inequality follows since either $a_j\geq\frac{1}{2}$, in which case $\frac{a_j}{1+a_j}\geq \frac{1}{2}$ or $0<a_j\leq \frac{1}{2}$, in which case $\frac{a_j}{1+a_j}\geq\frac{a_j}{2}$.
A: Hint: Since $1 + a_j \to 1$, we have that $1 + a_j < 2$ eventually. Then ponder the fact that
$$\frac{a_j}{1 + a_j} > \frac{a_j}{2}$$
A: Following the hint: If $a_j\gt 0$ has limit $0$, then for large enough $j$, we have $0 \lt a_j\le 1$. Thus for large enough $j$, we have $\frac{a_j}{a_j+1}\ge \frac{a_j}{2}$. Thus by Comparison $\sum_{j=1}^\infty \frac{a_j}{a_j+1}$ diverges. 
Added: We prove a detail that was left out in the hint. Suppose that $\sum \frac{a_j}{a+a_j}$ converges. Then $\lim_{j\to\infty} a_j=0$. For suppose this is not the case. Then there is an $\epsilon\gt 0$ such that $a_j\gt \epsilon$ for arbitrarily large $j$. If for a particular $j$, we have $a_j\ge 1$, then $\frac{a_j}{1+a_j}\ge \frac{1}{2}$. If $\epsilon \lt a_j\lt 1$, then $\frac{a_j}{1+a_j} \gt \frac{\epsilon}{2}$.  
