Definition:Convergent Product
Convergent Product in a Standard Number Field
Let $\mathbb K$ be one of the standard number fields $\Q, \R, \C$.
Nonzero Sequence
Let $\sequence {a_n}$ be a sequence of nonzero elements of $\mathbb K$.
Then:
- The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent
- its sequence of partial products converges to a nonzero limit $a \in \mathbb K \setminus \set 0$.
Arbitrary Sequence
Let $\sequence {a_n}$ be a sequence of elements of $\mathbb K$.
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent if and only if:
- there exists $n_0 \in \N$ such that the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some $b \in \mathbb K \setminus \set 0$.
The sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty a_n$ is then convergent to some $a \in \mathbb K$.
Convergent Product in Arbitrary Field
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Nonzero Sequence
Let $\sequence {a_n}$ be a sequence of nonzero elements of $\mathbb K$.
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent if and only if its sequence of partial products converges to a nonzero limit $a \in \mathbb K \setminus \set 0$.
Arbitrary Sequence
Let $\sequence {a_n}$ be a sequence of elements of $\mathbb K$.
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent if and only if:
- there exists $n_0 \in \N$ such that the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some $b \in \mathbb K \setminus \set 0$.
The sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty a_n$ is then convergent to some $a \in \mathbb K$.
Convergent Product in Normed Algebra
Let $\mathbb K$ be a division ring with norm $\norm {\,\cdot\,}_{\mathbb K}$.
Let $\struct {A, \norm {\,\cdot\,} }$ be an associative normed unital algebra over $\mathbb K$.
Let $\sequence {a_n}$ be a sequence in $A$.
Definition 1
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent if and only if there exists $n_0\in\N$ such that:
- $(1): \quad a_n$ is invertible for $n \ge n_0$
- $(2): \quad$ the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some invertible $b\in A^\times$.
Definition 2: for complete algebras
Let $\struct {A, \norm{\,\cdot\,} }$ be complete.
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ is convergent if and only if there exists $n_0 \in \N$ such that:
- the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some invertible $a\in A^\times$.
Divergent Product
An infinite product which is not convergent is divergent.
Divergence to zero
If either:
- there exist infinitely many $n \in \N$ with $a_n = 0$
- there exists $n_0 \in \N$ with $a_n \ne 0$ for all $n > n_0$ and the sequence of partial products of $\ds \prod_{n \mathop = n_0 + 1}^\infty a_n$ converges to $0$
the product diverges to $0$, and we assign the value:
- $\ds \prod_{n \mathop = 1}^\infty a_n = 0$
Also defined as
Some authors define an infinite product to be convergent if and only if the sequence of partial products converges.
The phrase:
- $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$
then becomes:
- $\ds \prod_{n \mathop = 1}^\infty a_n$ converges to $0$.
The definition used here has many desirable consequences, such as:
- Factors in Convergent Product Converge to One
- Convergence of Infinite Product Does not Depend on Finite Number of Factors
- Product of Convergent and Divergent Product is Divergent
which creates an analogy with series.