Limit of an infinite sum can be computed from limit of each component?

Question

Let $\{a_{n,k}\}_{n\in \mathbb{N}}$ be a sequence for some $k\in \mathbb{N}$. Suppose that $$ \lim_{n\rightarrow \infty} a_{n,k}=0 \text{ } \text{ } \forall k=1,...,2n $$ Does this imply $$ \lim_{n\rightarrow \infty} \sum_{k=1}^{2n} a_{n,k}=0 $$ ? Could you help me to understand why yes or no?

Answer 1

Suppose that $a_{n, k} = 1/n$. Then, for any fixed $k$, we have $a_{n, k} \to 0$ as $n \to \infty$. But, $$ \sum_{k=1}^{2n}a_{n, k} = \sum_{k=1}^{2n}\frac{1}{n} = \frac{2n}{n} = 2$$ and hence $$ \lim_{n \to \infty} \sum_{k=1}^{2n} a_{n, k} = 2$$