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For an infinite product $\prod a_k$ to converge we need

  1. at most finitely many zero factor, let be $m$ the maximum index of them
  2. $c=\lim_{n\to \infty}\prod_{k=m+1}^n a_k$ must exists, and
  3. $c\ne 0$.

My question is "Why the additional condition 3?"

Consider $$\tag{1} \prod_{k=1}^\infty \frac{n}{n+1}=\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots $$ The $n$th partial product would be $1/n$, thus the limit is zero. The definition above excludes (1) from the converging infinite products, but I do not understand what is bad about (1) converging to zero. There must be some consideration behind it.

EDIT My question could be read as follows: What are the advantages of this definition? Is there a better (easier to develop by excluding the zero) definition ? Why is the zero limit excluded (even in case there are no zero factors)?

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  • $\begingroup$ Where is this definition coming from ? Did you reproduce faithfully ? $\prod_{k=0}^\infty 0$ obviously converges. By the way, there is a typo in 2. $\endgroup$ – Yves Daoust Jan 18 at 10:58
  • $\begingroup$ The definition (for my understanding) is from Knopp, Theory and Application of Infinite Series, p. 218, archive.org/details/theoryandapplica031692mbp/page/n232 $\endgroup$ – Mohamed Ali Jan 18 at 11:08
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    $\begingroup$ My personal understanding is that when you take the logarithm of an infinite product converging stricto sensu, it turns to a converging series. Products that tend to zero are simply less interesting. $\endgroup$ – Yves Daoust Jan 18 at 11:25
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    $\begingroup$ @Peter It appears to me that Bronstein has the same definition: Ein unendliches Produkt heißt genau dann konvergent, wenn entweder (1.186) mit $b\ne 0$ vorliegt, oder diese Situation lässt sich nach Weglassen von endlich vielen Faktoren erreichen, die alle gleich null sind. Anderenfalls heißt das Produkt divergent. $\endgroup$ – Mohamed Ali Jan 18 at 11:31
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    $\begingroup$ and the following paragraph is "Ein konvergentes unendliches Produkt ist genau dann gleich null, wenn ein Faktor gleich null ist.", and that is exacly what @JackM referred to $\endgroup$ – Mohamed Ali Jan 18 at 11:32
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Yes, this is the definition in all books covering infinite products. We say $\prod \frac{n}{n+1}$ "diverges to $0$", and is not included when we say an infinite product "converges". The reason for this definition is that it is useful, for example in complex analysis.

One example: (there are many others) $$ \sin z= z \prod_{n=1}^\infty \left(1-{\frac {z^{2} } {\pi^2n^{2} } } \right) $$ where (for all complex $z$) it is a convergent infinite product. Therefore, we may read the zeros of $\sin$ from it directly. With infinite products that possibly diverge to $0$, we cannot do that.

As all mathematics students know, the principle $ab = 0 \Longrightarrow (a=0\text{ or }b=0)$ is very useful. We want to keep this useful fact for infinite products also! The is the first (and simplest, of many) reasons that convergence of infinite products is defined this way.

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I would like to add a footnote of Pringsheim from 1888, I think the first "complete" work on the topic. He discuss exactly this question and he gives, 132 years ago, the same arguments as @GEdgar: convenience (Bequemlichkeit) and conserving the properties of a finite product.

Pringsheim: Über die Konvergenz unendlicher Produkte. Mathem. Ann. Bd. 33, S. 119-154. 1888.

The book can be found https://archive.org/details/mathematischean46behngoog/page/n126

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