Does $\lim_{\lambda\to\infty} \sum_{n=1}^{\infty} a_n f_n(x)=\sum_{n=1}^{\infty} \lim_{\lambda\to\infty} a_n f_n(x)$? Does $\lim_{x\to\infty} \sum_{n=1}^{\infty} a_n f_n(x)=\sum_{n=1}^{\infty} \lim_{x \to\infty} a_n f_n(x)$?
Lets assume that $\lim_{x\to \infty} f_n(x)$ exists for each $n$, and that $\sum_{n=1}^{\infty} a_n f_n(x)$ converges uniformly on $\mathbb{R}$.
The reason I ask this question is because of the thread: Calculate $\lim_{|\lambda| \to +\infty} \int_{\mathbb{R}}f(x)|\sin(\lambda x)|dx $ Specifically Tolaso's answer.
This answer asserts that $\lim_{\lambda \to \infty}\sum_{n=1}^{\infty} \int_{a}^b \frac{(-1)^{n+1}}{4n^2-1}f(x)\cos(2n\lambda x)dx=0$ by the Riemann-Lebesgue Lemma, but as user Bungo points out, this requires the interchanging summation and limit.
One idea is to think along the lines of Abel's theorem, which states that if $\sum a_n x^n$ converges in $(-1,1)$ and if $\sum a_n$ converges, then $\lim_{x\to 1} a_n x^n=\sum a_n$.
Any comments and incomplete answers welcomed!
 A: Let $(a_{n})$ be a sequence of real numbers and $(f_{n})$ a sequence
of real-valued functions defined on $\mathbb{R}$. Suppose that:
(1) For each $n$, $f_n(+\infty):=\lim_{x\rightarrow+\infty}f_n(x)\in\mathbb{R}$
exists, and
$(2)$ $\sum_{n=1}^{\infty}a_{n}f_{n}(x)$ converges uniformly over
$\mathbb{R}$.
Then $\sum_{n=1}^{\infty}a_{n}f_{n}(+\infty)$ converges and $\sum_{n=1}^{\infty}a_{n}f_{n}(+\infty)=\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}a_{n}f_{n}(x)$.
Proof: Denote $S_{n}(x)=\sum_{k=1}^{n}a_{k}f_{k}(x)$, $x\in(-\infty, +\infty]$. Firstly, we
show that $\sum_{n=1}^{\infty}a_{n}f_{n}(+\infty)$ converges. Let
$\varepsilon>0$ be arbitrary. By assumption (2), there exists $N\in\mathbb{N}$
such that $|S_{n+p}(x)-S_{n}(x)|<\varepsilon$ whenever $n\geq N$,
$p\in\mathbb{N}$, $x\in\mathbb{R}$. Let $n\geq N$ and $p\in\mathbb{N}$
be arbitrary. Letting $x\rightarrow+\infty$, then we have: $|S_{n+p}(+\infty)-S_{n}(+\infty)|\leq\varepsilon$.
This shows that $\left(S_{n}(+\infty)\right)_{n}$ is a Cauchy-sequence
and hence it is convergent. That is, $\sum_{n=1}^{\infty}a_{n}f_{n}(+\infty)$
converges.
Next, we go to prove that $\lim_{x\rightarrow+\infty}\sum_{n=1}^{\infty}a_{n}f_{n}(x)=\sum_{n=1}^{\infty}\lim_{x\rightarrow+\infty}a_{n}f_{n}(x)$. Let $S(x)=\sum_{n=1}^{\infty}a_{n}f_{n}(x)$, $x\in\mathbb{R}$. Let
$l=\sum_{n=1}^{\infty}a_{n}f_{n}(+\infty)$. Let $\varepsilon>0$
be arbitrary. Choose $N_{1}$ such that $|S(x)-S_{n}(x)|<\varepsilon$
whenever $n\geq N_{1}$ and $x\in\mathbb{R}$. Recall that $S_{n}(+\infty)\rightarrow l$ as $n\rightarrow\infty$,
so there exists $N_{2}$ such that $|S_{n}(+\infty)-l|<\varepsilon$
whenever $n\geq N_{2}$. Let $N=\max(N_{1},N_{2})$. Since $\lim_{x\rightarrow+\infty}S_{N}(x)=S_{N}(+\infty)$,
there exists $x_{0}\in\mathbb{R}$ such that $|S_{N}(x)-S_{N}(+\infty)|<\varepsilon$
whenever $x\in[x_{0},\infty)$. Finally, let $x\in[x_0,\infty)$ be arbitrary.
We have: 
\begin{eqnarray*}
|S(x)-l| & \leq & |S(x)-S_{N}(x)|+|S_{N}(x)-S_{N}(+\infty)|+|S_{N}(+\infty)-l|\\
 & < & \varepsilon+\varepsilon+\varepsilon.
\end{eqnarray*}
This shows that $\lim_{x\rightarrow+\infty}\sum_{n=1}^{\infty}a_{n}f_{n}(x)=\sum_{n=1}^{\infty}\lim_{x\rightarrow+\infty}a_{n}f_{n}(x)$.
