Which functions map nonnegative convergent series to convergent series? Inspired by another question about The set of functions which map convergent series to convergent series (in the domain of the function), I ask the same question but about nonnegative series, which is more interesting I think, since there is at least one sufficient condition: $\lim_{x\rightarrow 0}\frac{f(x)}{x}$ exist and finite.
What is the general condition for a function $f:\mathbb{R^*}\rightarrow\mathbb{R}$ to satisfy:
If $\sum (x_n)_{n\in\mathbb{N}}$ is a convergent nonnegative series and $\forall n:x_n\in\operatorname{Domain}(f)$ then $\sum(f(x_n))_{n\in\mathbb{N}}$ is a convergent series.
 A: Call $f\colon (0,\infty)\to\mathbb R$ positive convergence preserving (hereafter $CP^+$) if for every convergent series $\sum x_n$ with all $x_n$ positive, the series  $\sum f(x_n)$ is also convergent.
Theorem. A function $f\colon (0,\infty)\to\mathbb R$ is $CP^+$ if and only if there exist $r>0$, $M>0$ such that $\frac{|f(x)|}{x}<M$ for all $x\in(0,r)$.
Proof.
First assume
that $\frac{|f(x)|}x<M$ for $x\in(0,r)$.
Let $\sum x_n$ be convergent with $x_n>0$ and let $\epsilon>0$ be given. 
Then there exists some $ n_0\in\mathbb N$ such that  $\sum_{k=n}^mx_n<\frac1M\epsilon$ whenever $m>n>n_0$.
Also from $x_n\to 0$ we conclude that $x_n<r$ and hence $|f(x_n)|<Mx_n$ for almost all $n$, say for all $n>n_1$. Then for all $m>n>\max\{n_0,n_1\}$,
$$ \left|\sum_{k=n}^mf(x_k)\right|\le \sum_{k=n}^mMx_k<\epsilon$$
and we conclude that $\sum f(x_n)$ converges. 
For the other direction, assume 
that $\frac{|f(x)|}x$ is not bounded on any interval $(0,r)$.
Then there is a sequence $x_n\to 0$ such that $\frac{|f(x_n)|}{x_n}\to \infty$. By picking a subsequence we can ensure that 


*

*$x_n<2^{-n}$ (so that $\sum x_n$ converges) and that 

*$|f(x_n)|>2^nx_n$.


Now consider a new series $$ \underbrace{x_1+x_1+\ldots +x_1}_{m_1}+\underbrace{x_2+x_2+\ldots +x_2}_{m_2}+\ldots$$
that repeats $x_n$ exactly $m_n:=\lceil\frac{2^{-n}}{x_n}\rceil$ times. Then the contribution of these summands is $m_nx_n<2^{-n}+x_n$ so that this new series still converges.
On the other hand, we see that  $|m_nf(x_n)|>\frac{2^{-n}}{x_n}\cdot 2^nx_n=1$, i.e. the image series fails to be Cauchy. $_\square$
