Let $\{x_k\}_{k\in\mathbb{N}}\subset\mathbb{R}_{>0},\ $ and for each $n\in\mathbb{N}:$ let $P(X_n)$ be the power set of $X_n = \{x_1,\ldots,x_n\} $ and then let $f_n:I_n\to P(X_n)$ be a bijection for some index set $I_n$ of $P(X_n).$
Proposition:
If $\displaystyle\sum_{n=1}^{\infty} x_n$ converges, then so does $\displaystyle\lim_{n\to\infty}\sum_{i\in I_n} \left( \prod_{x_j\in f_n(i)} x_j \right).$
Hopefully this notation makes sense.
Is the proposition true? And what about if we let $\{x_k\}_{k\in\mathbb{N}}\subset\mathbb{R},\ $ i.e. we allow $\displaystyle\sum_{n=1}^{\infty} x_n$ to be conditionally convergent also?
$$ \sum_{i\in I_n} \left( \prod_{x_j\in f_n(i)} x_j \right) = 1 + \sum_{k=1}^n x_k + \sum_{1 \le k < l \le n} x_k x_l + \cdots + \prod_{j=1}^n x_j \\ = (1 + x_1)(1+x_2) \cdots (1+x_n) \, , $$ so your question is equivalent to asking if the convergence of the series $\sum_{n=1}^{\infty} x_n$ implies that of the infinite product $\prod_{n=1}^{\infty} (1+x_n)$.
And that is indeed true if all $x_n \ge 0$: $$ \prod_{n=1}^{N} (1+x_n) \le \exp \left( \sum_{n=1}^{N} x_n\right) \, . $$ The sequence on the right is convergent and therefore bounded, so that the sequence on the left is increasing and bounded above.
Actually in this case the convergence of $\sum_{n=1}^{\infty} x_n$ is equivalent to the convergence of $\prod_{n=1}^{\infty} (1+x_n)$ to a non-zero value. See https://en.wikipedia.org/wiki/Infinite_product for more information.
For the case of a conditional convergent series, see for example Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, then $\prod_{n=1}^{\infty} (1+a_n)$ converges conditionally or diverges to 0..