Absolutely Convergent Product is Convergent

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Theorem

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

Let $\mathbb K$ be complete.

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


Then it is convergent.


Proof

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

We show that $\left\langle{P_n}\right\rangle$ is Cauchy.

We have, for $m > n$:

\(\displaystyle \norm{P_m - P_n}\) \(=\) \(\displaystyle \prod_{k \mathop = 1}^n \norm{1 + a_k} \cdot \norm{\prod_{k \mathop = n + 1}^m \paren{1 + a_k} - 1}\)
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle \) \(\le\) \(\displaystyle \prod_{k \mathop = 1}^n \paren{1 + \norm{a_k} } \paren{\prod_{k \mathop = n + 1}^m \paren{1 + \norm{a_k} } - 1}\)
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle \) \(=\) \(\displaystyle Q_m - Q_n\)


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.

$\blacksquare$