Equivalence of Definitions of Absolute Convergence of Product

Theorem
Let $\mathbb K$ be a field with absolute value $\left\vert{\cdot}\right\vert$.

Let $(a_n)$ be a sequence in $\mathbb K$.

Then the following definitions of absolute convergence of a product are equivalent:

1 implies 2
By the Monotone Convergence Theorem, it suffices to show that the partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ are bounded.

Because $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \left\vert{a_n}\right\vert}\right)$ converges, its partial products are bounded.

The result now follows from $\displaystyle \sum_{n \mathop = 1}^N|a_n| \leq \prod_{n \mathop = 1}^N\left({1 + \left\vert{a_n}\right\vert}\right)$.

Proof 1
By the Monotone Convergence Theorem, it suffices to show that the partial products of $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \left\vert{a_n}\right\vert}\right)$ are bounded.

By Exponential of x not less than 1+x:
 * $\displaystyle \prod_{n \mathop = 1}^N\left({1 + \left\vert{a_n}\right\vert}\right) \leq \prod_{n \mathop = 1}^N\exp(|a_n|) = \exp\left( \sum_{n \mathop = 1}^N|a_n|\right)$

Because $\displaystyle \sum_{n \mathop = 1}^\infty|a_n|$ converges, its partial sums are bounded.

Proof 2
By the Monotone Convergence Theorem, it suffices to show that the partial products of $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \left\vert{a_n}\right\vert}\right)$ are bounded.

By the AM-GM Inequality:
 * $\displaystyle \prod_{n \mathop = 1}^N\left({1 + \left\vert{a_n}\right\vert}\right) \leq \left(\frac{N + \sum_{n \mathop = 1}^N|a_n|}N \right)^N\leq \left(1 + \frac MN\right)^N$

where $M>0$ is such that $\displaystyle \sum_{n \mathop = 1}^N|a_n|\leq M$ for all $N$.

By definition of the real exponential, $\displaystyle\left(1 + \frac MN\right)^N \to \exp(M)$ as $N\to\infty$.

By Convergent Sequence in Metric Space is Bounded, $\left(1 + \frac MN\right)^N$ is bounded.

Also see

 * Equivalence of Definitions of Absolute Convergence of Product of Complex Numbers