(related to Proposition: Quotient of Convergent Complex Sequences)
We have to prove the proposition only for the special case of a constant complex sequence \((a_n)_{n\in\mathbb N}\) with \(a_n=1\) for all \(n\in\mathbb N\), because the general case follows then immediately from the corresponding proposition about the product of convergent complex sequences and from the identity \[\frac {a_n}{b_n}=a_n\cdot\frac 1{b_n}.\]
By hypothesis, the limit of the convergent complex sequences \((b_n)_{n\in\mathbb N}\) does not equal the complex zero (\(b\neq 0\)). Thus, we can find some index \(N(b)\) such that
\[|b_n - b| < \frac{|b|}{2}\quad\quad\forall n \ge N(b).\]
Therefore, it follows1 that \(b_n\neq 0\) for all \(n \ge N(b)\), because \(|b_n|\ge |b|/2\) for all \(n \ge N(b)\), or \(1/|b_n|\le 2/|b|\). We will need this estimation a little bit later.
Now, it follows from the convergence of \((b_n)_{n\in\mathbb N}\) that for a given \(\epsilon > 0\), there exists an index \(N(\epsilon)\in\mathbb N\) such that
\[|b_n - b |< \frac{\epsilon|b|^2}{2}\quad\quad\forall n\ge N(\epsilon).\]
It follows for all \(n\ge M(\epsilon,b):=\max(N(b),N(\epsilon))\) that
\[\left|\frac 1{b_n} - \frac 1b \right| = \left|\frac{ b - b_n}{b_n b}\right|=\frac{1}{|b_n||b|}\cdot |b_n - b| < \frac{2}{|b|^2}\cdot\frac{\epsilon|b|^2}{2}=\epsilon\quad\quad\forall n\ge M(\epsilon,b).\]
Thus, we have proven that :
\[\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} \frac{a_n}{b_n}=\frac{\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} a_n}{\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} b_n}.\]
This is just because each sequence member \(b_n\) is "closer" to \(b\) then it is to \(0\), interpreting \(|b|/2\) as half the distance from \(b\) to \(0\). ↩
