Product of Convergent Products is Convergent

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ converge to $a$.

Let $\displaystyle \prod_{n \mathop = 1}^\infty b_n$ converge to $b$.


Then $\displaystyle \prod_{n \mathop = 1}^\infty a_nb_n$ converges to $ab$.


Proof

Let $n_0 \in \N$ such that $a_n \ne 0$ for $n> n_0$.

Let $n_1 \in \N$ such that $b_n \ne 0$ for $n> n_1$.

Then $a_n b_n \ne 0$ for $n > n_2 = \max \set {n_0, n_1}$.

Let $p_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2 + 1}^\infty a_n$.

Let $q_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2 + 1}^\infty b_n$.

Then $p_n q_n$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2+1}^\infty a_n b_n$.

Because $p_n$ and $q_n$ converge to a nonzero limit, so does $p_n q_n$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n b_n$ converges.


Let $P_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$.

Let $Q_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty b_n$.

Then $P_n Q_n$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n b_n$.

By Limit of Product of Sequences, $P_n Q_n\to ab$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_nb_n$ converges to $a b$.

$\blacksquare$


Also see