Convergence of double sum implies convergence of inner sums

Question

Suppose $a_i,b_{ij}\in\mathbb{R}$ and $$ 0\le \lim_{n\to\infty} \sum_{i=1}^n\sum_{j=1}^n c_{ij}a_ia_j< \infty. $$

Does it follow that $\sum_{j=1}^\infty c_{ij}a_j$ converges for all $i$? (Clarification: The individual summands are not nonnegative, and I do not want to assume absolute summability.)

Answer 1

Try $c_{ii} = 2$ for all $i$, $c_{i1} = c_{1i} = -1$ for $i \ge 2$, all other $c_{ij} = 0$, and all $a_i = 1$. Then $\sum_{i=1}^n \sum_{j=1}^n c_{ij} a_i a_j = 2$ for all $n$ but $\sum_{j=1}^\infty c_{1j} a_j$ diverges.

Answer 2

A counter-example is $a_i=0$ for all $i$ and $b_{ij}=1$ for all $i,j$.

However, if $a_i \neq 0$ for all $i$ then your double sum doesn't make sense unless $\sum_j b_{ij}$ converges for all $i$.

Answer for the edited version: let $a_i=0$ for $n$ odd and $1$ for $n$ even. Then the hypothesis says $0 \leq \sum_i\sum_j c_{(2i)(2j)} <\infty$. This does not say anything about $c_{ij}$ for $i$ odd. So we cannot conclude that $\sum_{ij} c_{ij} a_j$ is convergent when $i$ is odd.