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I am confronted with infinite products. I've tried to read several textbooks on this subject, but most advanced math texts give it short shrift or else delve into its complexities. My question is basic.

Why is the infinite product typically written, using sigma notation, as $1 + a_{k}$. I get it has to do with the fact that convergence demands it approach the multiplicative inverse (which is $1$). Ok. But I don't really really get that. Is it because the $a_{k}$ is actually approaching zero, so you need "$1 + $" to get to "$1$" as the term approaches zero.

In other words, is there a basic for dummies explanation of why you have "$1 + $" in your expression?

Thanks so much.

anna
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Abcderia
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    Complex analysis textbooks are the best source for this kind of thing; formalizing the notion of infinite products was a huge part of the development of the subject. I would think of it as a change of units. A sum is not going to converge unless things are getting close to $0$, which requires no adjustment. A product is not going to converge unless things are getting pretty close to $1$, and for this reason many people consider the things being producted as $1 + (\text{whatever})$ to focus on the thing that's making the expression get close to $1$. Which will be a thing going to zero. – leslie townes Feb 17 '21 at 17:06
  • @leslietownes: Thanks so much. Helpful. – Abcderia Feb 17 '21 at 17:14
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    @DMcMor: here's just one university worksheet I found online. You'll see all the problems are "1 + ..." which is quite typical. http://oldwww.ma.man.ac.uk/~mdc/MATH41022/Background/Background%20Inf%20Prod.pdf – Abcderia Feb 17 '21 at 17:14
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    Absolute convergence of $\prod(1+a_n)$ reduces to absolute convergence of $\sum a_n$. https://encyclopediaofmath.org/wiki/Infinite_product#:~:text=A%20necessary%20and%20sufficient%20condition,if%20it%20is%20absolutely%20convergent. If the product is absolutely convergent, we can rearrange the factors. I answered a question that depended on this the other day. https://math.stackexchange.com/questions/4017806/an-infinite-product-for-frac-pi2/4017868#4017868 – saulspatz Feb 17 '21 at 17:17

2 Answers2

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Let $(z_k)_{k \ge 1}$ be a sequence of non-zero complex numbers and define $$p_n = \prod_{k = 1}^nz_k$$ for $n \ge 1$.

Let us say that $$\prod_{k = 1}^\infty z_k$$ converges if $(p_n)_{n \ge 1}$ converges to a non-zero complex number.

The benefit of this definition now is that we have the usual similarity with series that the product converges only if $z_k \to 1$. Now, the point is that instead of looking at $(z_k)$, we can always consider $w_k:= z_k - 1$. In this case, the product becomes $$\prod_{k = 1}^\infty (1 + w_k)$$ and that converges only if $w_k \to 0$. This is even more familiar with what we see in the case of series.

Aryaman Maithani
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For $a_n \gt 0$: $$\prod\limits_{n=1}^{\infty}p_n = \prod\limits_{n=1}^{\infty}(1+a_n) \lt \infty \Leftrightarrow \sum\limits_{n=1}^{\infty}a_n\lt \infty$$ Now all appropriate knowledge for series can be used for products.

zkutch
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