Suppose there exists an integer \(N\) such that \(0≤ a_n ≤ b_n\) for all \(n≥N\), and suppose that \(\sum^∞_{n=1} b_n\) converges. Let's define \(s_n\), \(t_n\) and \(t\):
Since both series have positive terms, then the sequences \(\{s_n\}\) and \(\{t_n \}\) are increasing. Also \(t_n \to t\), so \(t_n \le t\) for all \(n\).
Since \(a_i \le b_i\), we have \(s_n \le t_n\). Thus \(s_n \le t \) for all \(n\). This means \(\{s_n\}\) is increasing and bounded above, and therefore converges by the monotone convergence theorem. Thus \(\sum^∞_{n=1} a_n\) converges.
Suppose there exists an integer \(N\) such that \(a_n \ge b_n \ge 0\) for all \(n≥N\), and suppose that \(\sum^∞_{n=1} b_n\) diverges. This means \(t_n \to \infty\) as \(n \to \infty\). If \(a_i \ge b_i\), then \(s_n \ge t_n\). Thus \(s_n \to \infty \) as \(n \to \infty\). Therefore \(\sum^∞_{n=1} a_n\) diverges.