Limit Summation interchanging Is there a theorem which says when we can interchange the limit and sum as follow:
$$\lim_{x\to \infty} \sum_{n=1}^{\infty}f(x,n)= \sum_{n=1}^{\infty}\lim_{x\to \infty}f(x,n)$$
Note: In my case the sum  $\sum_{n=1}^{\infty}f(x,n)$ is finite at each finite $x\in \mathbb R$.
 A: Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality
$$ 
\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))
$$
holds.
This theorem can be found in books of Zorich (Mathematical Analysis II p. 381).
In this case we have: $T=\mathbb{N}\cup\{\infty\}$, $X=\mathbb{R}$, $\mathcal{B}_{T}=\mathcal{B}_{\mathbb{N}\cup\{\infty\}}$ the set of parts of $\mathbb{N}\cup\{\infty\}$, $\mathbb{B}_X=\mbox{ topoloy of metric } d(u,v)= |u-v|$. And $F_t(x)=\Sigma_{n=1}^t f(n,x)$. Then
$$ 
\lim_{x\to \infty}\lim_{t\to \infty}\Sigma_{n=1}^t f(n,x)
=
\lim_{t\to \infty}\lim_{x\to \infty}\Sigma_{n=1}^t f(n,x)
$$
A sufficient condition for uniform convergence of the limit
 $$
\lim_{x\to\infty}F(x,t)
 $$
 over $ t$ is the Cauchy  criterium
$$
\lim_{\quad x\to+\infty}\sup_{t}|F_t(x)-\lim_{x\to \infty}\lim_{t\to \infty}F_t(x)|=0
$$
