$\sum_{k=1}^\infty a_k$ converges iff $\sum_{k=1}^\infty a_k/(1+a_k)$ converges I just came across the following exercise in Schinazi, R. (2012). From Calculus to Analysis (1st ed., p. 54). Basel: Birkhäuser.

Let $a_n>0$. Show that $\sum_{k=1}^\infty a_k$ converges if and only if $$\sum_{k=1}^\infty \frac{a_k}{1+a_k}$$ converges.

Just from the looks of it I assume I'd need to use the comparison test for the $\Rightarrow$ direction. What about the other direction though?
 A: If $\sum\frac{a_k}{1+a_k}$ converges, then necessarily
$$
\frac{a_k}{1+a_k}\to0\text{ as }k\to\infty,
$$
in turn implying that $a_k\to0$ as $k\to\infty$.  Then $a_k\leq 1$ for $k$ sufficiently large, and therefore
$$
\frac{a_k}{1+a_k}\geq\frac{a_k}{2}\text{ for $k$ sufficiently large}.
$$
Then, use the fact that $\sum a_k$ converges if and only if $\sum\frac{a_k}{2}$ converges.
A: For the reverse implication, show that if $\sum a_n$ diverges then so does $\sum a_n/(1+a_n)$.
If $a_n$ is bounded and $a_n < B $ for all $n$ then $a_n/(1+a_n) > a_n/(1+B)$ and we have divergence by the comparison test.  
If $a_n >0$ is unbounded then there is a subsequence $a_{n_k} \to \infty$ and $a_{n_k}/(1 + a_{n_k}) \to 1$ implying divergence by the term test.
A: $\displaystyle \sum \dfrac{a_k}{1+a_k}$ converges 
$\rightarrow $
$\lim_{k \rightarrow \infty} \dfrac{a_k}{1+a_k}=0,$
$\rightarrow$
$a_k$ is bounded , see note.
There is a $n_0$ such that $ a_k \lt 1.$
$a_k= \dfrac{a_k(1+a_k)}{1+a_k} \lt \dfrac{a_k(1+1)}{1+a_k} =$
$\dfrac{2a_k}{1+a_k}.$
Comparison test.
Note:
$\lim_{k \rightarrow \infty} \dfrac{a_k}{1+a_k} = 0.$
For  $1 > \epsilon >0$ there is a $n_0$ such that
for $k\ge n_0$ : 
$\dfrac{a_k}{1+a_k} \lt \epsilon$, or
$a_k(1-\epsilon) \lt \epsilon ,$
$a_k \lt \dfrac{\epsilon}{1-\epsilon}.$
Choose $\epsilon =1/2 $ to get $a_k\lt 1.$
A: We will use the limit comparison test. $$\lim_{n\to \infty}\frac{a_n (1-a_n)}{a_n} = 1 > 0$$
The results follows 
