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If we know the following convergence criteria for infinite products is correct

$$\boxed{\sum a_{n}\text{ converges }\Leftrightarrow\prod\left(1+a_{n}\right)\text{ converges, for real }a_{n}>0}$$

Then it immediately follows that

$$\boxed{\sum|a_{n}|\text{ converges }\Leftrightarrow\prod(1+|a_{n}|)\text{ converges}}$$

because $|a_n|>0$, even for complex $a_n$ as required by the first criteria.

Question: This this logic correct? I am asking because I have seen source which derive the first criterion (eg), and then derive the second in a complicated way. What am I missing?

Penelope
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  • Well, if $a>0$ then $|a|=a$ – Rhys Hughes Aug 08 '21 at 17:54
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    Kind of, so long as your original sequence $\alpha$ contains no $0$ terms. Essentially, you you have performed universal instantiation on the original claim with the sequence $\beta: \mathbb N \to \mathbb R_{>0 } := \left| \alpha\right|$ for any given sequence $\alpha: \mathbb N \to \mathbb C \setminus {0}$.

    If your sequence does contain $0$-terms then removing them does not change the convergence of either series or product, so then you reduce to the prior case and then what you whish to prove holds.

     – user2628206 Aug 08 '21 at 17:56
  • hi @user2628206 thanks, this is helpful. In this case the factors in the product can't be 0 because $a_n>0$ is asserted. Have I misunderstood your point about zeros? – Penelope Aug 08 '21 at 20:26
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    On my reading you only asserted that $\forall n \in \mathbb N_: \alpha_n > 0$ for the statement in the first box. The claim $\forall n \in \mathbb N_: \alpha_n > 0$ doesn't makes sense in the complex case as $\mathbb C$ is not ordered. However if you implicitly had that $\alpha: \mathbb N \to \mathbb C$ contains no nonzero terms then my discussion about $0$-terms is superfluous. Note It's true that $\left|\alpha_n\right| > 0$ for $\alpha_n \in \mathbb C$ if and only if $\alpha_n \neq 0$. You didn't include any commentary about this case nor did you rule it out explicitly so I mentioned it. – user2628206 Aug 08 '21 at 20:33
  • i.e. just doing universal instantiation on the first statement will get you the claim $\boxed{\sum|a_{n}|\text{ converges }\Leftrightarrow\prod(1+|a_{n}|)\text{ converges} \text{ for real } \left|\alpha_n\right| > 0}$. This statement is strictly weaker than your second statement, as means one cannot deduce the claim for sequence $\alpha: \mathbb N \to \mathbb C$ where there is an $n$ such that $\alpha_n = 0$. However as I argued prior as $0$ terms do not change the values of either the left or the right hand side, nor the convergence, so it is a quick argument to remove this extra hypothesis. – user2628206 Aug 08 '21 at 20:40
  • You could call me out and say that I am being pernickety, and I wouldn't say you were incorrect. My argument though is that your second claim isn't immediate as you have to do a very small amount of reasoning. However your claim is spot on. Hope this helps – user2628206 Aug 08 '21 at 20:45
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    hi @user2628206 - I think I finally understand - my apologies, I'm not trained as a mathematician. You're right, the second statement is missing a condition $|a_n|>0$ which is equivalent to $a_n\neq0$ for complex $a_n$. – Penelope Aug 08 '21 at 22:14
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    Does this answer your question? $\prod (1+a_n)$ converges iff $ \sum a_n$ converges – amWhy Aug 12 '21 at 19:13

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