The set of functions which map convergent series to convergent series Suppose $f$ is some real function with the above property, i.e.
if $\sum\limits_{n = 0}^\infty {x_n}$ converges, then $\sum\limits_{n = 0}^\infty {f(x_n)}$ also converges. 
My question is: can anything interesting be said regarding the behavior of such a function close to $0$, other than the fact that $f(0)=0$?

Answer to the next question: no.
Let $f\colon\mathbb{R}\to\mathbb{R}$ be defined by
$$
f(x)=\begin{cases}
n\,x & \text{if } x=2^{-n}, n\in\mathbb{N},\\
x & \text{otherwise.}
\end{cases}
$$
Then $\lim_{x\to0}f(x)=f(0)=0$, $f$ transforms convergent series in convergent series, but $f(x)/x$ is not bounded in any open set containing $0$. In particular $f$ is not differentiable at $x=0$. This example can be modified to make $f$ continuous.
Proof. 
Let $\sum_{k=1}^\infty x_k$ be a convergent series. Let $I=\{k\in\mathbb{N}:x_k=2^{-n}\text{ for some }n\in\mathbb{N}\}$. For each $k\in I$ let $n_k\in\mathbb{N}$ be such that $x_k=2^{-n_k}$. Then
$$
\sum_{k=1}^\infty f(x_k)=\sum_{k\in I} n_k\,2^{-n_k}+\sum_{n\not\in I} x_n.
$$
The series $\sum_{k\in I} n_k\,2^{-n_k}$ is convergent. It is enough to show that also $\sum_{n\not\in I} x_n$ is convergent. This follows from the equality
$$
\sum_{n=1}^\infty x_n=\sum_{n\in I} x_n+\sum_{n\not\in I} x_n
$$
and the fact that $\sum_{n=1}^\infty x_n$ is convergent and $\sum_{k\in I} x_n$ absolutely convergent.
The proof is wrong. $\sum_{k\in I} x_n$ may be divergent. Consider the series
$$
\frac12-\frac12+\frac14-\frac14+\frac14-\frac14+\frac18-\frac18+\frac18-\frac18+\frac18-\frac18+\frac18-\frac18+\dots
$$
It is convergent, since its partial sums are
$$
\frac12,0,\frac14,0,\frac14,0,\frac18,0,\frac18,0,\frac18,0,\frac18,0,\dots
$$
The transformed series is
$$
\frac12-\frac12+\frac24-\frac14+\frac24-\frac14+\frac38-\frac18+\frac38-\frac18+\frac38-\frac18+\frac38-\frac18+\dots
$$
whose partial sums are
$$
\frac12,0,\frac12,\frac14,\frac34,\frac12,\frac78,\frac34,\frac98,1,\frac{11}8,\frac54,\dots
$$
which grow without bound.

On the other hand, $f(x)=O(x)$, the condition in Antonio Vargas' comment, is not enough when one considers series of arbitrary sign. Let
$$
f(x)=\begin{cases}
x\cos\dfrac{\pi}{x} & \text{if } x\ne0,\\
0 & \text{if } x=0,
\end{cases}
\quad\text{so that }|f(x)|\le|x|.
$$
Let $x_n=\dfrac{(-1)^n}{n}$. Then $\sum_{n=1}^\infty x_n$ converges, but
$$
\sum_{n=1}^\infty f(x_n)=\sum_{n=1}^\infty\frac1n
$$
diverges.
