nLab infinite product

Infinite products




Infinite products


An infinite product is a sequence of numbers (usually real or complex (a k) k(a_k)_{k\in\mathbb{N}} written as k=0 a k\prod_{k=0}^\infty a_k. Like an infinite series, we are interested in knowing whether such a product converges, and if so, what it converges to.

As an operator

For any group or semigroup GG, we can inductively define the finitary product \mathbb{N}-action

n=0 ()()(n):(G)×G\product_{n=0}^{(-)} (-)(n) : (\mathbb{N} \to G) \times \mathbb{N} \to G


n=0 0a(n)=a(0)\product_{n=0}^{0} a(n) = a(0)
n=0 k+1a(n)=a(k+1) n=0 ka(n)\product_{n=0}^{k+1} a(n) = a(k+1) \product_{n=0}^{k} a(n)

for all k:k:\mathbb{N} a:Ga:\mathbb{N} \to G, such that currying the action results in the infinite product operator

n=0 ()(n):(G)(G)\prod_{n=0}^\infty (-)(n) : (\mathbb{N} \to G) \to (\mathbb{N} \to G)

over the function G G -module G\mathbb{N} \to G.


In a cartesian closed category CC, let (G,)(G,\cdot) be an multiplicative magma object, let (N,0,s)(N,0,s) be a natural numbers object in CC, and let a:NGa:N\to G be an infinite sequence object of elements in GG. Then there exists an infinite sequence object b:NGb:N\to G called the left infinite product object of aa inductively defined by b(0)=a(0)b(0) = a(0) and b(s(n))=b(n)a(s(n))b(s(n)) = b(n) \cdot a(s(n)), and an infinite sequence object c:NGc:N\to G called the right infinite product object of aa inductively defined by c(0)=a(0)c(0) = a(0) and c(s(n))=a(s(n))c(n)c(s(n)) = a(s(n)) \cdot c(n). The element b(n)b(n) is called the left nn-th partial product of the sequence aa, and the element c(n)c(n) is called the right nn-th partial product of the sequence aa. If the magma object is commutative and associative, then the left and right infinite product objects of aa are equal and just called an infinite product object d= n:Na(n)d = \prod_{n:N} a(n) of aa, where the element d(n)d(n) is the nn-th partial product.


A naive definition of convergence, by analogy with the sum of a series, would be that k=0a k=limN k=0 Na k\underoverset{k=0}{\infty}{\prod} a_k = \underset{N\to\infty}{\lim} \prod_{k=0}^N a_k if the latter limit exists. However, this has the flaw that it could happen that this limit exists and yet limka k\underset{k\to\infty}{\lim} a_k might not, whereas we would like to be able to say that if k=0 a k\prod_{k=0}^\infty a_k converges then limka k=1\underset{k\to\infty}{\lim} a_k = 1 (by analogy with the fact that if k=0 a k\sum_{k=0}^\infty a_k converges then limka k=0\underset{k\to\infty}{\lim} a_k = 0). This failure can happen for two reasons:

  1. If some a k=0a_k = 0, then limN k=0 Na k=0\underset{N\to\infty}{\lim} \prod_{k=0}^N a_k = 0 since the partial products are eventually 00, regardless of the eventual behavior of the sequence (a k)(a_k).

  2. If |a k|M<1{\vert a_k\vert}\le M\lt 1, then limN k=0 Na k=0\underset{N\to\infty}{\lim} \prod_{k=0}^N a_k = 0, whereas the sequence (a k)(a_k) might approach any limit of absolute value <1\lt 1 or have no limit at all.

To avoid these “pathological” situations, we make the following modified definition.


Suppose at most finitely many of the a ka_k are zero. We say that k=0 a k\prod_{k=0}^\infty a_k converges if

limN k=1a k0 a k \underset{N\to\infty}{\lim} \prod_{k=1\quad a_k\neq 0}^\infty a_k

exists and is nonzero. If this is the case, we say that

k=0 a k={0 if some a k=0 the above limit otherwise \prod_{k=0}^\infty a_k = \begin{cases} 0 &\quad \text{if some }a_k = 0\\ \text{the above limit} &\quad\text{otherwise} \end{cases}

If the above limit equals 0, one sometimes says that k=0 a k\prod_{k=0}^\infty a_k diverges to 0.


Last revised on May 28, 2021 at 22:07:51. See the history of this page for a list of all contributions to it.