Absolutely Convergent Product is Convergent

Theorem

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.

Let $\mathbb K$ be complete.

Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.

Then it is convergent.

Proof

Let $P_n$ and $Q_n$ denote the $n$th partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ and $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ respectively.

We show that $\sequence {P_n}$ is Cauchy.

We have, for $m > n$:

\(\ds \norm {P_m - P_n}\)

\(=\)

\(\ds \prod_{k \mathop = 1}^n \norm {1 + a_k} \cdot \norm {\prod_{k \mathop = n + 1}^m \paren {1 + a_k} - 1}\)

\(\ds \)

\(\)

\(\ds \)

\(\ds \)

\(\le\)

\(\ds \prod_{k \mathop = 1}^n \paren {1 + \norm {a_k} } \paren {\prod_{k \mathop = n + 1}^m \paren {1 + \norm {a_k} } - 1}\)

\(\ds \)

\(\)

\(\ds \)

\(\ds \)

\(=\)

\(\ds Q_m - Q_n\)

Because $\sequence {Q_n}$ converges, $\sequence {Q_n}$ is Cauchy.

By the above inequality, $\sequence {P_n}$ is Cauchy.

Because $\mathbb K$ is complete, $\sequence {P_n}$ converges to some $a \in \mathbb K$.

By Absolutely Convergent Product Does not Diverge to Zero, the product converges.

$\blacksquare$