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 $f(0)=0$?
 A: If $f$ is not continuous at $0$, then we can find a sequence $x_n$ that converges to $0$ but $f(x_n)$ doesn't converge to $0$. First get a subsequence $y_n$ of $x_n$ with $|f( y_n)| > r$ for some $r>0$. Next choose some subsequence $z_n$ of $y_n$ so that
$\sum z_n$ converges. However the series $\sum f(z_n)$ diverges and it follows that $f$ is continuous at $0$.
A: I'm quite late on this one, but I think the result is nice enough to be included here.
Definition A function $f : \mathbb R \to \mathbb R$ is said to be convergence-preserving (hereafter CP) if $\sum f(a_n)$ converges for every convergent series $\sum a_n$.
Theorem (Wildenberg): The CP functions are exactly the ones which are linear on some neighbourhood of $0$.
Proof (Smith): Clearly, whether $f$ is CP only depends on the restriction of $f$ on an arbitrary small neighbourhood of $0$. Since the linear functions are CP, the condition is clearly sufficient. Let's prove that it is also necessary.
We will prove two preliminary results.
Lemma 1: $f$ CP $\Rightarrow$ $f$ continuous at $0$. 
Proof: Let's suppose that $f$ isn't continuous at 0. This implies that there exists a sequence $\epsilon_n \to 0$ and a positive real $\eta > 0$ such that $\forall n, |f(\epsilon_n)| \geq \eta$. But it is easy to extract a subsequence $\epsilon_{\phi(n)}$ such that $\sum \epsilon_{\phi(n)}$ converges (take $\phi$ such that $\epsilon_{\phi(n)} \leq 2^{-n}$, for instance). For such a subsequence, we still have that $|f(\epsilon_{\phi(n)})| \geq \eta$. This prevents $\sum f(\epsilon_{\phi(n)})$ to converge and, thus, $f$ to be CP, a contradiction.
Lemma 2: The function $(x, y) \mapsto f(x+y) + f(-x) + f(-y)$ vanishes on some neighbourhood of $0$.
Proof: If it didn't, one would be able to find sequences $x_n \to 0$ and $y_n \to 0$ s.t. $\forall n, f(x_n + y_n) + f(-x_n) + f(-y_n) \neq 0$. Up to some extraction, we can assume that $\delta_n = f(x_n + y_n) + f(-x_n) + f(-y_n)$ always has the same sign (let's say $\delta_n > 0$, for the sake of simplicity.)
Consider now the series $$\begin{array}{l@{}l}
(x_0 + y_0) &+ (-x_0) + (-y_0) +  \cdots + (x_0 + y_0) + (-x_0) + (-y_0)\\ 
& +(x_1 + y_1) + (-x_1) + (-y_1) +  \cdots + (x_1 + y_1) + (-x_1) + (-y_1)\\
&+\cdots\\
&+(x_n + y_n) + (-x_n) + (-y_n) +  \cdots + (x_n + y_n) + (-x_n) + (-y_n)+\cdots,
\end{array}$$
where every triplet of termes $(x_i+y_i) + (-x_i) + (-y_i)$ is repeated $M_i > 0$ times, for some integer $M_i > 0$.
Because $x_n \to 0$ and $y_n \to 0$ and the three terms $x_i + y_i, -x_i, -y_i$ add to 0, it is easy to see that this series is convergent, regardless of the choice of the $M_i$'s.
On the other hand, if we choose $M_i \geq \delta_i^{-1}$, the image of our series by $f$ is $$\begin{array}{l@{}l}
f(x_0 + y_0) &+ f(-x_0) + f(-y_0) +  \cdots + f(x_0 + y_0) + f(-x_0) + f(-y_0)\\ 
& +f(x_1 + y_1) + f(-x_1) + f(-y_1) +  \cdots + f(x_1 + y_1) + f(-x_1) + f(-y_1)\\
&+\cdots\\
&+f(x_n + y_n) + f(-x_n) + f(-y_n) +  \cdots + f(x_n + y_n) + f(-x_n) + f(-y_n)+\cdots,
\end{array}$$
which diverges, for every line adds to $M_i \delta_i > 1$. Again, this in direct contradiction with the CPness of $f$.
If we apply the result of lemma 2 with $y = 0$, we get that $f(-x) = -f(x)$. So we can rewrite lemma 2 in the following way:
$\exists \eta > 0 : \forall x, y \in (-\eta, \eta), f(x+y) = f(x) + f(y)$.
This property and the continuity at 0 imply first the continuity on the whole of $(-\eta, \eta)$ and it is then not hard to adapt the classical proof to show that $f$ is linear on $(-\eta, \eta)$. Q.E.D. 
A: 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.
