Prove that if $a_n\gt 0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ diverges. Prove that if $a_n\gt 0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ diverges. 

This is the solution to this problem, but I'm having a hard time understanding it. Why does $a_k/(1+a_k)$ not converge to $0$ if $a_k$ doesn't converge to $0$? 
I'd appreciate it if anyone could answer this question for me.
 A: The solution is flawed. The conclusion in the final sentence does not follow from the preceding line (an upper bound won't do; we need a lower bound). But if we're assuming $0<a_k<1$, then $\dfrac{a_k}{1+a_k} > \dfrac{a_k}2$, and so we have bounded our series below by a divergent series.
The solution presented heads off in a far more complicated direction than is needed and then doesn't finish it off correctly. 
By the way, if $a_k\ge 1$, then $\dfrac{a_k}{1+a_k}\ge \dfrac12$, and so infinitely many such terms will clearly give a divergent subseries.
A: I think there's just a typo in the solution; the final inequality should be $>$, not $<$. From the solution you gave, we have:
$$a_k - \frac{a_k}{1+a_k} < \frac{a_k}2$$
Now move the $\frac{a_k}{1+a_k}$ term to the right-hand side and the $a_k$ to the left to get:
$$\frac{a_k}2 < \frac{a_k}{1+a_k}$$
Taking partial sums, the left side is unbounded, therefore so is the right.
A: If $ v_k=\dfrac{a_k}{1 +a_k} $ then $a_k=\dfrac {v_k }{1-v_k } $
Then if $v_k $ converge to zero , $ a_k $ converge also to zero.
