(related to Proposition: Product of Convegent Complex Sequences)

Let the complex sequences \((a_n)_{n\in\mathbb N}\) and \((b_n)_{n\in\mathbb N}\) be convergent with \(\lim_{n\rightarrow\infty} a_n=a\) and \(\lim_{n\rightarrow\infty} b_n=b\). It follows from the corresponding proposition that they are bounded by some positive real number \(B > 0\). By making \(B\) sufficiently large, without any loss of generality, we can assume \(|a_n|\le B\) and \(|b_n| \le B\) for all \(n\in\mathbb N\). Further, let \(\epsilon > 0\). It follows that

  1. there is an \(N(\epsilon,B)\in\mathbb N\) with \(|a_n - a| < \frac{\epsilon}{2B}\) for all \(n > N(\epsilon,B)\), and that
  2. there is an \(M(\epsilon,B)\in\mathbb N\) with \(|b_n - b| < \frac{\epsilon}{2B}\) for all \(n > M(\epsilon,B)\).

Therefore, for all \(n > \max(N(\epsilon,B),M(\epsilon,B))\) we get by virtue of the triangle inequality. \[|a_nb_n - ab| = |a_n(b_n - b)+ (a_n-a)b| \le |a_n||b_n - b| + |a_n - a||b| < B\frac{\epsilon}{2B} + \frac{\epsilon}{2B}B=\epsilon.\]

Since \(\epsilon\) can be arbitrarily small chosen, it follows that

\[\lim_{n\rightarrow\infty} (a_n \cdot b_n)=\lim_{n\rightarrow\infty} a_n \cdot \lim_{n\rightarrow\infty} b_n.\]

Thank you to the contributors under CC BY-SA 4.0!




  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983