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Infinite series (sums) are discussed a lot, but infinite products less so. Despite having tried hard to find reading material on infinite products, I have only found lecture notes and not proper texts.

Having read these, eg "Background Notes Theory of Infinite Products", I am puzzled as to why infinite products that do have a limit of 0 are said to "diverge".

Here is one example:

$$ \prod_{n=2}^{\infty}(1-\frac{1}{n}) $$

As usual we take the partial sum and consider the limit.

$$ \require{cancel} \begin{align} \prod_{n=2}^{N}(1-\frac{1}{n}) &= \prod_{n=2}^{N}(\frac{n-1}{n}) \\ \\ &= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{4} \times \ldots \times \frac{N-1}{N} \\ \\ &= \frac{1}{N} \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to 0.

Because this limit is finite, albeit 0, I feel this should be considered convergent. What key insight am I missing?

Update: according to lecture notes, "diverge to zero" means there are an infinite number of factors that are zero, so this specific example doesn't "diverge to zero".

Qiaochu Yuan
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Penelope
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1 Answers1

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The simplest way to explain this convention is that it ensures that an infinite product $\prod r_i$ of positive reals converges iff its logarithm $\sum \log r_i$ converges in the usual sense, which reduces the theory of infinite products (of positive reals) to the theory of infinite sums. If an infinite product diverges to $0$ then its logarithm diverges to $-\infty$. The divergence of the infinite product you describe turns out to be equivalent to the divergence of the harmonic series, for example (Example 5 in the notes you've linked).

This convention also has the desirable property that it makes convergence closed under taking inverses: $\prod r_i$ converges to $\ell$ iff $\prod r_i^{-1}$ converges to $\ell^{-1}$.

More abstractly, you might say that when we deal with infinite products we restrict our attention to the positive reals $\mathbb{R}_{+}$, because taking products involving $0$ is not very interesting. So $0$ is excluded and takes on the role of a "point at infinity" (and again this can be motivated by thinking about the logarithm $\log : \mathbb{R}_{+} \to \mathbb{R}$).

(Considering the nonzero reals is not much of an upgrade over the positive reals, because for convergence only finitely many terms can be negative anyway. There are reasons to consider the nonzero complex numbers though, e.g. when considering the Euler product of the Riemann zeta function for complex values of $s$.)

Qiaochu Yuan
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  • Please could you say a few words about why my interpretation of those notes is not quite correct? I am keen to understand. – Penelope Dec 27 '20 at 19:29
  • @Tariq: oh, my apologies, I have actually misread the notes. I thought your update was correct but it is incorrect; your example is in fact an example of an infinite product that diverges to $0$. It's the multiplicative analog of the harmonic series, and can even be used to prove that the harmonic series diverges. – Qiaochu Yuan Dec 27 '20 at 19:32
  • interestingly there is a difference on opinion. In these different notes (http://www.math.ualberta.ca/~isaac/math324/s12/zeta.pdf) there is a statement on page 2 "If infinitely many factors an are zero, then we say the product diverges to zero" which is different to wikipedia which says. – Penelope Dec 27 '20 at 19:34
  • Yeah, I'm not aware of this convention being universal. There are several different ways you could set up the theory. It depends on what you want to do. – Qiaochu Yuan Dec 27 '20 at 19:36
  • this comment is also relevant ... https://math.stackexchange.com/a/778372/319008 – Penelope Dec 27 '20 at 22:44