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 $\ds \sum_{n \mathop = 1}^\infty a_n$ are bounded.

Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ converges, its partial products are bounded.

By Bounds for Finite Product of Real Numbers:
 * $\ds \sum_{n \mathop = 1}^N \norm {a_n} \le \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} }$

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

By Bounds for Finite Product of Real Numbers:
 * $\ds \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} } \le \map \exp {\sum_{n \mathop = 1}^N \norm {a_n} }$

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

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

By the AM-GM Inequality:
 * $\ds \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} } \le \paren {\frac {N + \sum_{n \mathop = 1}^N \norm {a_n} } N }^N \le \paren {1 + \frac M N}^N$

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

By definition of the real exponential, $\paren {1 + \dfrac M N}^N \to \map \exp M$ as $N \to \infty$.

By Convergent Sequence in Metric Space is Bounded, $\paren {1 + \dfrac M N}^N$ is bounded.

Also see

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