Absolutely Convergent Product Does not Diverge to Zero/Proof 1

Theorem

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

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

Then it is not divergent to $0$.

Proof

By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$.

Let $n_1 \ge n_0$.

Aiming for a contradiction, suppose the product diverges to $0$.

Then:

$\ds \prod_{n \mathop = n_1}^\infty \paren {1 + a_n} = 0$

By Norm of Limit:

$\ds \prod_{n \mathop = n_1}^\infty \norm {1 + a_n} = 0$

By the Triangle Inequality and Squeeze Theorem:

$\ds \prod_{n \mathop = n_1}^\infty \paren {1 - \norm {a_n} } = 0$

By the Weierstrass Product Inequality, we have for $N \ge n_1$:

$\ds \prod_{n \mathop = n_1}^N \paren {1 - \norm {a_n} } \ge 1 - \sum_{n \mathop = n_1}^N \norm{a_n}$

Taking limits:

$0 \ge 1 - \ds \sum_{n \mathop = n_1}^\infty \norm {a_n}$

Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent:

$\ds \sum_{n \mathop = n_1}^\infty \norm {a_n} < 1$

for $n_1$ sufficiently large.

This is a contradiction.

$\blacksquare$