# Factors in Absolutely Convergent Product Converge to One

## 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:

$a_n \to 0$

## Proof 1

We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.

By the definition of absolutely convergent product, $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent.

 $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ $=$ $\ds \paren {1 + \size {a_1} } \paren {1 + \size {a_2} } \paren {1 + \size {a_3} } \paren {1 + \size {a_4} } \cdots$ expanding out the product $\ds$ $=$ $\ds 1 + \paren {\size {a_1} + \size {a_2} + \size {a_3} + \size {a_4} + \cdots} + \paren{\size {a_1} \size {a_2} + \size {a_1} \size {a_3} + \size {a_1} \size {a_4} + \cdots + \size {a_2} \size {a_3} + \size {a_2} \size {a_4} + \cdots} + \paren {\size {a_1} \size {a_2} \size {a_3} + \size {a_1} \size {a_2} \size {a_4} + \cdots + \size {a_2} \size {a_3} \size {a_4} + \cdots} + \cdots$ $\ds$ $=$ $\ds 1 + \sum_{i \mathop = 1}^\infty \size {a_i} + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty \size {a_i} \size { a_{i + j} } + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \size {a_i} \size { a_{i + j} } \size {a_{i + j + k} } + \cdots$

From the above, we see that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} } > \sum_{i \mathop = 1}^\infty \size {a_i}$

And since $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent, then $\ds \sum_{i \mathop = 1}^\infty \size {a_i}$ is convergent

Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.

$a_n \to 0$

$\blacksquare$

## Proof 2

We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.

Let $b_n = \paren {1 + a_n}$

Then $\ds \prod_{n \mathop = 1}^\infty b_n$ is absolutely convergent.

From Absolutely Convergent Product is Convergent, $\ds \prod_{n \mathop = 1}^\infty b_n$ is convergent.

$b_n \to 1$

Thus:

$a_n \to 0$

$\blacksquare$