Proving the convergence of $\bigg (\sum_{n}^{\nu} \bigg( \sum_{n}^{\chi} |\log(p_{n})+\log(q_{n})| \bigg ) \bigg )$?

Question

In the text "A Collection of Problems On Complex Analysis" by L.I. Volkovyskii, G.L. Lunts, and I.G Aramanovich.

The infinite products $\prod_{n=1}p_{n}$ and $\prod_{n=1}q_{n}$ converge. Investigate the convergence of the infinite product in $(1)$

$$\prod_{}^{}(p_{n}+q_{n})\tag{1}$$

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Is my proof to $(1)$ that follows, valid ?

$\text{Discussion}$

One can suppose that $\prod_{}^{}(p_{n}+q_{n})$ converges for values in $\mathbb{C}$ in other words that the bound

$$\prod_{}^{}(p_{n}+q_{n}) < \infty.$$

Holds for values of $\mathbb{C}$ , and also note that $p_{n}, q_{n} \in \mathbb{C}$

$\text{Lemma (1.2)}$

The product $\prod_{}(1+a_n)$ converges absolutely iff $\prod_{}(1+|a_n|)<\infty$, which only occurs when iff $\sum_{}^{}|a_n|<\infty.$

$\text{Lemma (1.3)}$

Recall the product in $(1)$ using the notion that one can convert from products to series where in $(2)$:

$(2)$

$$\log \prod s_n = \sum \log s_n.$$

Using $(2)$ one can arrive at the following in $(3)$

$(3)$

$$\log \prod_{n}^{}(p_{n}+q_{n}) = \sum |\log(p_{n}) + \log(q_{n})| = \bigg (\sum_{n}^{\nu} \bigg (\sum_{n}^{\chi} |\log(p_{n})+\log(q_{n}) | \bigg ).$$

$\text{Lemma (1.4)}$

From $(3)$ one has to show in $(4)$

$(4)$

$$ (s_{\nu \chi})^{\infty}_{\nu,\chi= 1}=\bigg (\sum_{n}^{\nu} \bigg( \sum_{n}^{\chi} |\log(p_{n})+\log(q_{n})| \bigg ) \bigg ) < \infty.$$

$\text{Lemma (1.5)}$

Taking $\lim_{\nu , \chi \rightarrow \infty } \bigg (\sum_{n}^{\nu} \bigg( \sum_{n}^{\chi} |\log(p_{n})+\log(q_{n})| \bigg ) \bigg )$ one achieves the following in $(5)$, since the double sequence of partial sums $(s_{\nu \chi})^{\infty}_{\nu,\chi= 1} < \infty.$

$(5)$

$$\lim_{\nu , \chi \rightarrow \infty }(s_{\nu \chi})^{\infty}_{\nu,\chi= 1}=\lim_{\nu,\chi \rightarrow \infty}\big (\sum_{n}^{\nu} \bigg(\sum_{n}^{\chi} |\log(p_{n})+\log(q_{n})| \big ) \big ) = \psi.$$

From $(5)$ the absolute convergence of $(\sum_{n}^{\chi} \big(\sum_{n}^{\nu } |\log(p_{n})+\log(q_{n})| \big ) \big )$ implies the convergence of $(\sum_{n}^{\chi} \big(\sum_{n}^{\nu } \log(p_{n})+\log(q_{n}) \big ) \big)$ which in turn shows the convergence of $\prod_{}^{}(p_{n}+q_{n}).$

Answer 1

Nothing can be said in general. Below is an example where the new product diverges.

Let $1/ 2 > \delta>0$ be small, and let $p_n = q_n = 1 + (-1)^n \delta.$ Note that $$\prod_{n \le k} p_n = (1 + \delta)^{o(k)} (1- \delta^2)^{ (k-o(k))/2}, \,\, o(k) := k \,\,\mathrm{mod}\, 2,$$ which is smaller than $2 (1 - \delta^2)^{k/2 -1},$ and the product decays to zero. But $$ \prod_{n \le k} (p_n + q_n) = (2 + 2\delta)^{o(k)} (2- 2 \delta^2)^{ (k-o(k))/2}, $$ and for $\delta < 1/2, $ this is $> (3/2)^{k/2} \uparrow \infty.$