# nLab infinite product

Infinite products

### Context

#### Algebra

higher algebra

universal algebra

# Infinite products

## Definition

An infinite product is a sequence of numbers (usually real or complex $(a_k)_{k\in\mathbb{N}}$ written as $\prod_{k=0}^\infty a_k$. 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 $G$, we can inductively define the finitary product $\mathbb{N}$-action

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

as

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

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

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

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

## Internalisation

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

## Convergence

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

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

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

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

###### Definition

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

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

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

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

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

## Examples

Last revised on May 28, 2021 at 18:07:51. See the history of this page for a list of all contributions to it.