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

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 $$(1)$$

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

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

• If $\prod_{n=1}p_n$ converges, doesn't that mean $p_n\rightarrow 1$? Feb 10, 2018 at 1:31
• @JohnBarber I don't think so this sort of problem is a "soft" one $p_{n}$ can be any value in $\mathbb{R}$ Feb 10, 2018 at 2:09
• I must be missing something here. I'm assuming these are infinite products, right? How can $\prod_{n=1}p_n$ and $\prod_{n=1}q_n$ converge unless $p_n\rightarrow 1$ and $q_n\rightarrow 1$? In which case $p_n + q_n\rightarrow 2$, and $\prod_{n=1}(p_n + q_n)$ must diverge. Feb 10, 2018 at 2:39
• Wait. This can work if the original products converge to zero, in which case $p_n$ and $q_n$ could converge to numbers less than 1. As long as $p_n + q_n$ converges to something less than or equal to 1, then the combined product might converge. Feb 10, 2018 at 2:43
• Putting $p_{n}, q_{n}$ in $\mathbb{R}$ was a huge mistake after looking at some examples it would be impossible to get these products to converge in $\mathbb{R}$ even with setting bounds on $p_{n}$ and $q_{n}$ so to address the issue I put $q_{n},p_{n} \in \mathbb{C}$. I'm sorry for the mistake :'>( !. Feb 10, 2018 at 3:09

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 $> (3/2)^{k/2} \uparrow \infty.$
• From reading your answer $p_n = q_n = 1 + (-1)^n \delta.$ is defined in $\mathbb{R}$ would the Theorem proposed be false in $\mathbb{R}$ but true in $\mathbb{C}$ ? Feb 13, 2018 at 19:43
• $\mathbb{R} \subset \mathbb{C},$ so no. Feb 13, 2018 at 20:13