Is it true that $\sum_k a_k(n)\to0$ as $n\to\infty$ if $a_k(n)\to0$ for all $k$ and $\sum_k a_k(n)$ converges for all $n$? Let $a_k(n)$ be such that $\sum_k a_k(n)$ converges for all $n$ and $a_k(n)\to0$ as $n\to\infty$ for all $k$. Then I think it holds that $\sum_k a_k(n)\to0$ as $n\to\infty$. 
For example, if $\displaystyle\sum_{k=1}^\infty a_k$ is any convergent series then $\displaystyle\sum_{k=1}^\infty a_k(n)=\sum_{k=1}^\infty \frac{a_k}{n}=\frac1n\sum_{k=1}^\infty a_k\to0.$
I guess one should write $a_k(n)=\delta(n)\varepsilon(k)$ where $\delta(n)$ and  $\varepsilon(k)$ respectively don't depend on $k$ and $n$... then one gets $\sum_k a_k(n)=\delta(n)\sum_k\varepsilon(k)\to0$ as $n\to\infty$, but how can we be sure that such $\delta$ and $\varepsilon$ exist?
 A: $1, 0,0,0,\dots$
$0,1, 0,0,0,\dots $
$0,0,1, 0,0,0,\dots $
$0,0,0,1, 0,0,0,\dots $
$\dots$
$\dots$
A: No. If I follow your question correctly, that is definitely not a valid conclusion. Here's an example: 
Define $a_k(n) = 0$ if $n \le k$, but $a_k(n) = \frac{1}{2^{n-k}}$ otherwise. 
Then clearly $a_k(n) \to 0$ for all $k$ as $n \to \infty$.
And the sum $\sum_k a_k(n)$ is just 1. 
Hence $\lim_{n \to \infty} \sum_k a_k(n)$ is $1$, not zero. 
A: Consider $a_k(n) = \begin{cases}2^{n-k} & \text{if} \ k \ge n \\ 0 & \text{if} \ k < n\end{cases}$. 
Then, for any $k$, we have $a_k(n) = 0$ for all $n > k$, and thus, $\displaystyle\lim_{n \to \infty}a_k(n) = 0$. 
Also, for any $n$, we have $\displaystyle\sum_{k = 0}^{\infty}a_k(n) = \sum_{k = 0}^{n-1}0 + \sum_{k = n}^{\infty}2^{n-k} = 0 + 2 = 2$. 
So, $\displaystyle\sum_{k = 0}^{\infty}a_k(n)$ converges for all $k$, but $\displaystyle\lim_{n \to \infty}\sum_{k = 0}^{\infty}a_k(n) = 2 \neq 0$. 
To make the conclusion hold, you will probably need some sort of monotonicity or uniformity condition.
A: If you consider $\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$, as $n$ gets large, the sequence $\langle\frac{1}{k^n}\rangle_{_n}\rightarrow0$ for all $k$ but the partial sums tend to $1$. While it may hold for certain sums it surely isn't applicable for all.
addendum: here I am taking $a_k(n)=\frac{1}{k^n}$
