Absolutely Convergent Product Does not Diverge to Zero

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Theorem

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

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


Then it is not divergent to $0$.


Proof 1

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

Let $n_1\geq n_0$.

Suppose that the product diverges to $0$.

Then $\displaystyle \prod_{n=n_1}^\infty(1+a_n) = 0$.

By Norm of Limit, $\displaystyle \prod_{n=n_1}^\infty \norm{1+a_n} = 0$.

By the Triangle Inequality and Squeeze Theorem, $\displaystyle \prod_{n=n_1}^\infty(1- \norm{a_n}) = 0$.

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

$\displaystyle \prod_{n=n_1}^N(1- \norm{a_n}) \geq 1 - \displaystyle \sum_{n=n_1}^N \norm{a_n}$

Taking limits, $0\geq 1 - \displaystyle \sum_{n=n_1}^\infty \norm{a_n}$.

Because $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ is absolutely convergent, $\displaystyle\sum_{n=n_1}^\infty \norm{a_n}<1$ for $n_1$ sufficiently large.

This is a contradiction.

$\blacksquare$


Proof 2

We have that $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 - \norm{a_n}}\right)$ is absolutely convergent.

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

Thus $\displaystyle \sum_{n \mathop = n_0}^\infty \log \left({1 - \norm{a_n}}\right)$ is absolutely convergent.

Suppose that the product diverges to $0$.

Then $\displaystyle \prod_{n=n_0}^\infty(1+a_n) = 0$.

By Norm of Limit, $\displaystyle \prod_{n=n_0}^\infty \norm{1+a_n} = 0$.

By the Triangle Inequality and Squeeze Theorem, $\displaystyle \prod_{n=n_0}^\infty(1- \norm{a_0}) = 0$.

By Logarithm of Infinite Product of Real Numbers, $\displaystyle \sum_{n \mathop = n_0}^\infty \log \left({1 - \norm{a_n}}\right)$ diverges to $-\infty$.

This is a contradiction.

$\blacksquare$


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