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 !!!
 A: A convergent example:
$$ \sum_{n=1}^{\infty} \left( \frac{1}{x+n} + \frac{1}{x-n} \right) = \pi\cot{\pi x}-\frac{1}{x}, $$
and the convergence is uniform on closed intervals not containing integer points.
Clearly every function in the sum tends to $0$ as $x \to \infty$, but the entire sum does not.

Edit: An even better example, perhaps: set
$$ s_n(x) = \chi_{[n,n+1)}(x), $$
i.e. $1$ for $n \leqslant x < n+1$, and $0$ otherwise. It is clear that $s_n(x) \to 0$ as $x \to \infty$, but it is also clear, since every real number is in precisely one interval of the form $[n,n+1)$, that
$$ \sum_{n=-\infty}^{\infty} s_n(x)=1. $$
A: This is not true. Consider $s_n(x) = 1/x$. We have $\lim_{x \to \infty} s_n(x) = 0 = t_n$. Clearly, $\sum_{n=1}^{\infty} t_n =0 < \infty$. However,
$$S_n(x) = \sum_{k=1}^n s_n(x) = \dfrac{n}x$$ and $\lim_{n \to \infty} S_n(x)$ doesn't exist.
A: As others have mentioned, this is not true in general.  But there is a very important case where it is true: power series.  Abel's theorem says that  if $S(x) = \sum_{n=0}^\infty a_n x^n$ and $S(1)$ converges, then $\lim_{x \to 1-} S(x) = S(1)$.
Since you're interested in $x \to \infty$, this will require a transformation to apply to your case.  Thus take some increasing function $g: [0,\infty) \to \mathbb R$ such that $\lim_{x \to \infty} g(x) = L > 0$ exists, and a series of the form
$S(x) = \sum_{n=0}^\infty a_n g(x)^n$.  If $S = \sum_{n=0}^\infty a_n L^n$ converges, then $\lim_{x \to \infty} S(x) = S$.
A: You can also find examples in which all series are convergent, yet you cannot pass to the limit. Try this: 
$$
s_n(x)=
\begin{cases}
1, & n=[x] \\
0, & \text{otherwise}
\end{cases}
$$
Here $[x]$ is the floor or ceiling function (choose the one you like most, it does not matter). 
To guarantee that you can pass to the limit under the sum you need some extra properties on the convergence of $s_n(x)$. 
A: No.
Let $s_n(x)=\frac 1x$, $s_n=0$. Then $S=\sum_{n=1}^\infty s_n$ trivially converges. However, $S(x)=\sum_{n=1}^\infty s_n(x)$ is clearly divergent, since the summand is a constant (with respect to $n$) greater than $0$.
A: 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.
