# 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 $x\in \mathbb R$.

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This is a general treatment of a connection between uniform convergence and the interchange of limits: math.la.asu.edu/~jss/courses/fall06/mat472/… – joriki Mar 23 '12 at 20:34

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.
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 $$\lim_{\quad x\to+\infty}\sup_{t}|F_t(x)-\lim_{x\to \infty}\lim_{t\to \infty}F_t(x)|=0$$