Factors in Absolutely Convergent Product Converge to One

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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.

By Terms in Convergent Series Converge to Zero:

$a_n \to 0$


$\blacksquare$


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.

By Factors in Convergent Product Converge to One:

$b_n \to 1$

Thus:

$a_n \to 0$

$\blacksquare$


Also see


There is believed to be a mistake here, possibly a typo.
In particular: I'm puzzled about this. This result, or at least Proof $2$, relies on Absolutely Convergent Product is Convergent, not the other way round.
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