Definition
  • We say that \(\lim\limits_{n\to\infty} x_n = \infty\) if for all \(M > 0\), we can find \(N \in \mathbb{N}\) such that whenever \(n > N\), \(x_n > M\).
  • We say that \(\lim\limits_{n\to\infty} x_n = -\infty\) if for all \(M > 0\), we can find \(N \in \mathbb{N}\) such that whenever \(n > N\), \(x_n < -M\).

Based on this definition, then

Divergence Theorem
Let \(x_n\) be a sequence such that \(\lim x_n = \infty\). Then
  1. If \(y_n\) is bounded below, then \(\lim x_n+y_n = \infty\).
  2. If \(y_n\) is bounded below by a postive \(M > 0\), then \(\lim x_ny_n = \infty\).

Proof

Let \(M > 0\) be given. Since \(y_n\) is bounded below, then there exists some \(m\) such that \(y_n \geq m\) for all \(n \in \mathbb{N}\). Now set

$$ \begin{align*} K = \max\{1, M-m\} \end{align*} $$

Then \(K > 0\). Moreover, since \(\lim x_n = \infty\), then by definition, for any positive number (set this number to \(K\)), there is an \(N \in \mathbb{N}\) such that for all \(n \geq N\),

$$ \begin{align*} x_n > K > M-m \end{align*} $$

Hence for all \(n \geq N\),

$$ \begin{align*} x_n + y_n &> x_n + m \\ &> K + m \\ &\geq (M - m) + m = M \end{align*} $$

Then, since \(M > 0\) is arbitrary, then by definition we have \(\lim (x_n+y_n) = \infty\) as we wanted to show. \(\ \blacksquare\)

References