Product of Convergent Products is Convergent

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

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

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

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

Proof
Let $n_0\in\N$ such that $a_n\neq0$ for $n> n_0$.

Let $n_1\in\N$ such that $b_n\neq0$ for $n> n_1$.

Then $a_nb_n\neq0$ for $n>n_2=\max(n_0,n_1)$.

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

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

Then $p_nq_n$ is the $n$th partial product of $\displaystyle \prod_{n=n_2+1}^\infty a_nb_n$.

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

Thus $\displaystyle \prod_{n=1}^\infty a_nb_n$ converges.

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

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

Then $P_nQ_n$ is the $n$th partial product of $\displaystyle \prod_{n=1}^\infty a_nb_n$.

By Limit of Product of Sequences, $P_nQ_n\to ab$.

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

Also see

 * Product of Convergent and Divergent Product is Divergent