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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?

Muzi
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    Is it $\sum |a_n|$ or $\sum a_n$ ? Is $a_n^2 = |a_n|^2$ ? – LL 3.14 Oct 12 '20 at 21:13
  • note that the answer in the linked question above treats only $k=2$ but the same ideas clearly work for higher $k$ since we truncate the Taylor series of the logarithm there and apply absolute convergence from then on, while the rest form a finite sum of convergent series – Conrad Oct 12 '20 at 21:47

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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 $\prod_n (1+a_n)$.

Andreas
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  • What is $m_i$? And isn't $n\to\infty$ with infinitly many constants $b_{j,h}$? – Its_me Oct 12 '20 at 21:53
  • this is not what the problem asks (finitely many consecutive power sums convergent and the next one absolutely convergent) – Conrad Oct 12 '20 at 21:53
  • and actually your statement is not correct, in other words even if of all $\sum a_n^k, k \ge 1$ converge it may happen that $\Pi(1+a_k)$ diverges as we can see by taking $a_n=e^{in}/\log n, b_n=e^{-in}/\log n, c_n=-a_n, d_n=-b_n, n \ge 2$ and the product $\Pi(1+a_n)(1+b_n)(1+c_n)(1+d_n)$ where we can for example index it modulo $4$ so each term is $(1+x_m)$ with $x_{4m+1}=a_{m}$ etc; it is very easy to see that $\sum x_m^k$ converges but $(1+a_n)(1+b_n)(1+c_n)(1+d_n)=1+q_n, q_n>0, \sum q_n =\infty$ so the product cannot converge – Conrad Oct 12 '20 at 22:14
  • Ok thanks to everyone. It wasn't clear to me, since the task description never mentions infinite sums. Concerning $m_i$: these are exponents, the full description is given in the quoted Wikipedia link for Newton's identities. – Andreas Oct 13 '20 at 12:13