If $k \in \mathbb N$ and if $\sum_n a_n, \sum_n a_n^2, \dots, \sum_n |a_n|^k$ converges then $\prod_n (1+a_n)$ converges Suppose that $(a_n)_n$ is a sequence of complex numbers with the property that $\sum_n a_n, \sum_n a_n^2, \dots,\sum_n |a_n|^k$ all converge where $k\in \mathbb N$. I want to show that this implies that the product $\prod_n (1+a_n)$ converges. I read this in a paper without proof so it should be true. Di you know how to prove this?
The sums $p_k = \sum_n a_n^k$ are known as power sums.
Using elementary symmetric polynomials $e_j$, you have (see here) that
$$
\prod_n (1+a_n) = \sum_{j=1}^n e_j
$$
Further, Newton's identities allow to represent all  elementary symmetric polynomials again as finite sums of products of power sums, see here. So we have that
$$
\prod_n (1+a_n) = \sum_{j=1}^n \sum_h b_{j,h} \prod_{i =1}^j c_{j,i} p_i^{m_i}
$$
with a finite number of fixed constants $b_{j,h}$ and $c_{j,i}$. Hence if all power sums converge, so does $\prod_n (1+a_n)$.
The sums $p_k = \sum_n a_n^k$ are known as power sums.
Using elementary symmetric polynomials $e_j$, you have (see here) that
$$
\prod_n (1+a_n) = \sum_{j=1}^n e_j
$$
Further, Newton's identities allow to represent all  elementary symmetric polynomials again as finite sums of products of power sums, see here. So we have that
$$
\prod_n (1+a_n) = \sum_{j=1}^n \sum_h b_{j,h} \prod_{i =1}^j c_{j,i} p_i^{m_i}
$$
with a finite number of fixed constants $b_{j,h}$ and $c_{j,i}$. Hence if all power sums converge, so does $\prod_n (1+a_n)$.