If each term in a sum converges, does the infinite sum converge too? Let $S(x) = \sum_{n=1}^\infty s_n(x)$ where the real valued terms satisfy
$s_n(x) \to s_n$ as $x \to \infty$ for each $n$. 
Suppose that $S=\sum_{n=1}^\infty s_n< \infty$.
Does it follow that $S(x) \to S$ as $x \to \infty$?
I really do not recall any theorems... please point out !!!

Here is another condition which is sufficient to allow you to interchange the limit and the sum.
Let
$$
  S(x) = \sum_{n=1}^\infty s_n(x),
  \qquad
  S_m = \sum_{n=1}^m \lim_{x \rightarrow \infty} s_n(x) = \sum_{n=1}^m s_n,
  \qquad
  S_m(x) = \sum_{n=1}^m s_n(x).
$$
We know that $S_m \rightarrow S$ and want to know if also $S(x) \rightarrow S$.
If each of your functions is bounded by some number,
$$
  | s_n(x) | \le g_n \qquad \forall n \in \mathbb N
$$
and the $(g_n)$ are summable, i.e.,
$$
  \sum_{n=1}^\infty g_n < \infty,
$$
then we have
$$
 |S(x) - S_m(x)| = \left| \sum_{n=m+1}^\infty s_n(x) \right| \le \sum_{n=m+1}^\infty g_n,
$$
and clearly the right-hand side converges to zero uniformly in $x$ (since it doesn't even depend on $x$). In other words,
$$
  S_m(x) \rightarrow S(x) \quad \text{uniformly in } x.
$$
Therefore, we are allowed to exchange the sum and the limit and obtain
$$
  \lim_{x \rightarrow \infty} S(x)
  = \lim_{x \rightarrow \infty} \lim_{m\rightarrow\infty} S_m(x)
  = \lim_{m\rightarrow\infty} \lim_{x \rightarrow \infty} S_m(x)
  = \lim_{m \rightarrow \infty} S_m.
$$
This is the principle of dominated convergence.
