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?
$1, 0,0,0,\dots$

$0,1, 0,0,0,\dots $

$0,0,1, 0,0,0,\dots $

$0,0,0,1, 0,0,0,\dots $

$\dots$

$\dots$
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 $a_k(n)=\frac{1}{k^n}$