= Solution
The family is a <subset> of all complex <sequences>, so its <cardinality> is at most
$$
|\mathbb C^{\mathbb N}|=\mathfrak c^{\aleph_0}=\mathfrak c.
$$
For each $z\in\mathbb C$, the <sequence> $(z,0,0,\ldots)$ belongs to the family because its displayed <series> has just one nonzero <first-order term>. These <sequences> are distinct as $z$ varies, giving a <lower bound> $\mathfrak c$. Thus
$$
\boxed{\text{cardinality}=\mathfrak c.}
$$
The source condition is <convergence> of $\sum_j a_j^2$, without absolute-value bars. The argument uses exactly that printed condition and makes no assertion that it is equivalent to square summability of complex <sequences>. The same <upper and lower bounds> would also work for the stronger absolute-square condition, but no alteration of the source is needed.
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