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I am revising Complex Analysis and I am a bit confused. I have a couple of results from lectures which say that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if the sum $\sum_{n=1}^\infty \log(1+a_n) $ converges absolutely.

And also, the infinite product converges absolutely if and only if the $|a_n|$ are summable.

Consider the example: $$\prod_{n=2}^\infty \left(1- \frac 1n\right) $$ It can easily be shown that the partial product $\prod_{n=2}^{N} (1- \frac 1n) = \frac 1N $ which tends to zero as $N\to \infty$

Maybe I'm interpreting the theorems wrong, but the sums:

$\sum_{n=2}^\infty |\log(1-\frac 1n)| $ and $\sum_{n=1}^\infty |-\frac 1n| $ both diverge, so from the results I listed at the start, the product cannot converge, but it does - to zero.

I'm getting quite confused here so I would appreciate some help!

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    $\begingroup$ It diverges to zero. $\endgroup$ – A little lime May 2 '14 at 11:09
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If partial products tends to zero as $N\to \infty$ we say that infinite product diverges to $0$.

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    $\begingroup$ @user141421 No kidding? For some easy exposition, see en.wikipedia.org/wiki/Infinite_product $\endgroup$ – Did May 2 '14 at 13:05
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    $\begingroup$ @user141421: My point was (although your reply to my question with other questions sort of ruined the process of getting there) that terms have context. In the context of products we say that something diverges to $0$, just like that in the context of the extended real line we say that something converges to $\infty$. $\endgroup$ – Asaf Karagila May 2 '14 at 14:19
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    $\begingroup$ @user141421: Again, we are talking about the context of products, in which we generally only talk about products of strictly positive numbers. If you wanna plug into the context things which are out of context, don't be surprised if the results are flustered mumbling and lack of definitive answers. $\endgroup$ – Asaf Karagila May 2 '14 at 14:24
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    $\begingroup$ @user141421 No you are not supposed to accept something just because it is written on WP. In the present case though, it is written on WP and it happens to be explained in every textbook on infinite series and infinite products. (Last comment from me on this matter, I have little patience for your type.) $\endgroup$ – Did May 2 '14 at 14:40
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    $\begingroup$ @Did Here is a proposition from Stein and Shakarchi. Page 141: "Proposition 3.1 If $\Sigma \vert a_n \vert < \infty$, then the product $\prod(1+a_n)$ converges. Moreover, the product converges to $0$ if and only if one of its factors is $0$." This never talks about convergence or divergence to zero, when the sum diverges. When the sum diverges, one needs to be careful as to what happens to the product and argue whether the product converges or diverges accordingly. Fanaticism in mathematics and sciences is worse than religious fanaticism. $\endgroup$ – user141421 May 2 '14 at 15:25
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The right answer to your question is that the statement should read as follows:

$\displaystyle\prod_{n=1}^\infty (1+a_n)$ converges to a non-zero real number if and only if the sum $\displaystyle\sum_{n=1}^\infty \log(1+a_n) $ converges absolutely.

As you have rightly observed the infinite product in your example does indeed converge to $0$. There are some people who would argue that the infinite product diverges to $0$, since the corresponding infinite sum diverges. However, this is incorrect, especially if you view the infinite product as $$\displaystyle\prod_{n=1}^\infty (1+a_n) = \lim_{N \to \infty} b_N, \text{ where } b_N = \displaystyle\prod_{n=1}^N (1+a_n)$$


EDIT Also, to emphasize the fact that converging to zero is the right term to use, I quote the following proposition from Stein and Shakarchi. Page 141:

"Proposition $3.1$: If $\Sigma \left\vert a_n \right\vert < \infty$, then the product $\prod(1+a_n)$ converges. Moreover, the product converges to $0$ if and only if one of its factors is $0$."

Hence, converges to zero is the right term to use. When the sum diverges, one needs to be careful as to what happens to the product and argue whether the product converges to zero or diverges accordingly.

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    $\begingroup$ No, the product diverges to $0$ because $0$ does not belong to the topological group $\mathbb{C}^\times$. Infinite products are like infinite sums, well, they are infinite sums in (abelian) topological groups where you write the operation as multiplication. We say $\prod u_n$ converges if it converges in $\mathbb{C}^\times$. Then we make special conventions for products containing zero factors. $\endgroup$ – Daniel Fischer May 2 '14 at 14:02
  • $\begingroup$ @DanielFischer Who cares? For this problem, I work over $\mathbb{R}$. $\endgroup$ – user141421 May 2 '14 at 14:14
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    $\begingroup$ @DanielFischer Here is a proposition from Stein and Shakarchi. Page 141: "Proposition 3.1 If $\Sigma \vert a_n \vert < \infty$, then the product $\prod(1+a_n)$ converges. Moreover, the product converges to $0$ if and only if one of its factors is $0$." This never talks about convergence or divergence to zero, when the sum diverges. When the sum diverges, one needs to be careful as to what happens to the product and argue whether the product converges or diverges accordingly. $\endgroup$ – user141421 May 2 '14 at 15:26

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