Given a sequence $\{a_{n,p}\}$ with $(n,p)\in\Bbb{Z}×\Bbb{N}$. Suppose $\lim\limits_{p\to+\infty}a_{n,p}=a_n\neq 0$ for each $n\in\Bbb{Z}$, $\sum\limits_{n=-\infty}^{+\infty}|a_n|<+\infty$ and $S=\sum\limits_{n=-\infty}^{+\infty}a_n$. Prove or disprove that: $$\lim_{p\to+\infty}\sum_{n=-p}^{p}a_{n,p}=S.$$
Any hints will welcome!!
On
The assertion is false. Find a counterexample such that for each $n$, the sequence $a_{n,p}$ tends to $a_n$ as $p\to\infty$. But set things so that the larger $|n|$ is, the longer $a_{n,p}$ stays big -- say $a_{n,p}$ remains $1$ whenever $p\le|n|$, and only after $p>|n|$ will you allow $a_{n,p}$ to take the constant value $a_n$. (You should visualize this with a picture.)
Take $a_{n,p} = \frac{1}{p} + \frac{1}{(1+|n|)^2}$, where $p \in \mathbb{N}$ and $n \in \mathbb{Z}$. Then $$ a_{n,p} \to \frac{1}{(1+|n|)^2} :=a_n, \text{ as } p\to \infty, $$ clearly $\sum_{n} |a_n| < \infty $, and let $S= \sum_{n\in \mathbb{Z}}{a_n}$. Then $$ \sum_{n = -p}^p a_{n,p} = \frac{2p+1}{p} + \sum_{n=-p}^p \frac{1}{(1+|n|)^2} \to 2 + S \text{ as } p \to \infty. $$