Equivalence of Definitions of Absolute Convergence of Product

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

Let $\sequence{a_n}$ be a sequence in $\mathbb K$.

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

By Bounds for Finite Product of Real Numbers, $\displaystyle \sum_{n \mathop = 1}^N \norm{a_n} \leq \prod_{n \mathop = 1}^N\left({1 + \norm{a_n}}\right)$.

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

By Bounds for Finite Product of Real Numbers:
 * $\displaystyle \prod_{n \mathop = 1}^N\left({1 + \norm{a_n}}\right) \leq \exp\left( \sum_{n \mathop = 1}^N \norm{a_n}\right)$

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

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

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

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

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