How do I prove that if $|a_{nk}| = o(b_{nk})$ as $n \to \infty$ for each $k$, then $\sum_{k=1}^n |a_{nk}| = o( \sum_{k=1}^n b_{nk} )$?
EDIT: David C. Ullrich showed this isn't true in general.
However, if we have uniform convergence of $|a_{nk}|/b_{nk}$ to zero, then we can conclude that for every $\epsilon > 0$, there is a $N$ high enough such that for $n > N$, $|a_{nk}| < \epsilon \times b_{nk}$ for every $k$.
Summing on each side,
$\sum_{k=1}^n|a_{nk}| < \epsilon\sum_{k=1}^n b_{nk} $
which shows that $\sum_{k=1}^n|a_{nk}| = o(\sum_{k=1}^n b_{nk})$
Counterexample:
Let $b_{nk}=2^{-k}$. Let $$a_{nk}=\begin{cases}1,&(n\le k^2), \\0,&(n>k^2).\end{cases}$$
Then $\sum_{k=1}^na_{nk}$ is roughly $n-\sqrt n$, wich tends to infinity, while $\sum_{k=1}^nb_{nk}$ is bounded.