Does divergence of $\sum a_k$ imply divergence of $\sum \frac{a_k}{1+a_k}$? Does divergence of $\sum a_k$ imply divergence of $\sum \frac{a_k}{1+a_k}$?
Note: $a_k > 0 $
I understand that looking at the contrapositive statement, we can say that the convergence of the latter sum implies  $\frac{a_k}{1+a_k}\rightarrow 0$ but from here is it possible to deduce that $a_k\rightarrow 0$ because it is not completely straightforward. If we assume $a_k$ to be convergent, this trivially follows but it could diverge in which case this is nontrivial to me.
 A: If $a_k\geq 0$ for every $k$, then you can reason as follows.
If $(a_k)$ does not converge to $0$, then there exists a subsequence $(a_{n_k})$ with limit $l\in (0,+\infty]$.
It is then easy to verify that
$$
\frac{a_{n_k}}{1+a_{n_k}} \to
\begin{cases}
1, & \text{if}\ l=+\infty,\\
\frac{l}{1+l}\neq 0, &\text{if}\ l\in (0,+\infty),
\end{cases}
$$
hence $\sum \frac{a_k}{1+a_k}$ cannot converge, so that diverges to $+\infty$.
Consider now the case $\lim_k a_k = 0$.
Then there exists $N\in\mathbb{N}$ such that $0\leq a_k \leq 1$ for every $k\geq N$, so that
$$
\frac{a_k}{1+a_k} \geq a_k/2 
\qquad \forall k\geq N,
$$
and again the series diverges by comparison.
Remark: the point is that
$$
\frac{a_k}{1+a_k} \geq
\begin{cases}
1/2, &\text{if}\ a_k > 1,\\
a_k / 2, & \text{if}\ a_k\in [0,1].
\end{cases}
$$
A: If $\sum_n \frac{a_n}{1+a_n}$ converges, $\frac{a_n}{1+a_n}$ goes to $0$ as $n$ goes to $\infty$, hence $a_n$ goes to $0$ as $n\to \infty$.
But since $\lim_{n\to \infty} \frac{a_n}{\frac{a_n}{1+a_n}} = 1$, for large enough $n$, $\left|\frac{a_n}{\frac{a_n}{1+a_n}} - 1 \right|\leq \frac 12$, hence $a_n\leq \frac 32 \frac{a_n}{1+a_n}$. Therefore, $\sum a_n$ converges by comparison.
A: In the case where the $a_n$ can change sign, let
$$b_n = \frac{(-1)^n}{\sqrt{n}}\quad \text{and}\quad a_n=\frac{b_n}{1-b_n} = \frac{(-1)^n}{\sqrt{n}}\frac{1}{1-\frac{(-1)^n}{\sqrt{n}}} =
\frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + O\left(\frac{1}{n^{3/2}}\right)$$
then one has $b_n = \frac{a_n}{1+a_n}$, the series $\sum a_n$ diverges but the series $\sum b_n$ converges.
