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. $_\square$
