Exercises §9 — Ross, Elementary Analysis, 2nd Edition
Exercise 9.9. Suppose there exists \(N_0\) such that \(s_n \le t_n\) for all \(n > N_0\).
Exercise 9.9. (a) Prove that if \(\lim s_n = +\infty\), then \(\lim t_n = +\infty\).
Proof. Let \(M > 0\). Then there exists \(N > N_0\) such that \(n > N\) implies \(s_n > M\). Since \(N > N_0\), this also implies \(t_n \ge s_n > M\). Thus \(\lim t_n = +\infty\). \(\square\)
Exercise 9.9. (b) Prove that if \(\lim t_n = -\infty\), then \(\lim s_n = -\infty\).
Proof. Let \(M < 0\). Then there exists \(N > N_0\) such that \(n > N\) implies \(t_n < M\). Since \(N > N_0\), this also implies \(s_n \le t_n < M\). Thus \(\lim s_n = -\infty\). \(\square\)
Exercise 9.9. (c) Prove that if \(\lim s_n\) and \(\lim t_n\) exist, then \(\lim s_n \le \lim t_n\).
Proof. Assume for contradiction that \(\lim s_n > \lim t_n\). Let \(L_1 = \lim s_n\) and \(L_2 = \lim t_n\), so \(L_1 > L_2\). Let \[ \varepsilon = \frac{L_1 - L_2}{2} > 0. \] Choose \(N_1 > N_0\) such that for all \(n > N_1\), \[ |s_n - L_1| < \varepsilon \;\Longrightarrow\; -\varepsilon < s_n - L_1 < \varepsilon \;\Longrightarrow\; s_n > L_1 - \varepsilon. \] Now for the same \(\varepsilon\) choose \(N_2 > N_0\) such that for all \(n > N_2\), \[ |t_n - L_2| < \varepsilon \;\Longrightarrow\; -\varepsilon < t_n - L_2 < \varepsilon \;\Longrightarrow\; t_n < L_2 + \varepsilon. \] Note that \[ L_1 - \varepsilon = \frac{2L_1 - L_1 + L_2}{2} = \frac{L_1 + L_2}{2} \;\Longrightarrow\; s_n > \frac{L_1 + L_2}{2}, \] \[ L_2 + \varepsilon = \frac{2L_2 + L_1 - L_2}{2} = \frac{L_1 + L_2}{2} \;\Longrightarrow\; t_n < \frac{L_1 + L_2}{2}. \] But this implies \(t_n < \frac{L_1 + L_2}{2} < s_n\) for all \(n > N = \max(N_1, N_2) > N_0\), which contradicts \(s_n \le t_n\) for all \(n > N > N_0\). Thus \(\lim s_n \le \lim t_n\). \(\square\)
Exercise 9.10.
Exercise 9.10. (a) Show that if \(\lim s_n = +\infty\) and \(k > 0\), then \(\lim(k s_n) = +\infty\).
Proof. Let \(M > 0\). Now let \(M_1 = \frac{M}{k} > 0\). There exists \(N\) such that for all \(n > N\), \(s_n > M_1 = \frac{M}{k} > 0\). This implies \(k s_n > M\), which means \(\lim(k s_n) = +\infty\). \(\square\)
Exercise 9.10. (b) Show that \(\lim s_n = +\infty\) if and only if \(\lim(-s_n) = -\infty\).
Proof.
(\(\Rightarrow\)) Suppose \(\lim s_n = +\infty\). Let \(M < 0\). Note \(-M > 0\). There exists \(N\) such that for all \(n > N\), \(s_n > -M \Rightarrow -s_n < M\), so \(\lim(-s_n) = -\infty\).
(\(\Leftarrow\)) Suppose \(\lim(-s_n) = -\infty\). Let \(M > 0\). Note \(-M < 0\). There exists \(N\) such that for all \(n > N\), \(-s_n < -M \Rightarrow s_n > M\), so \(\lim s_n = +\infty\). \(\square\)
Exercise 9.10. (c) Show that if \(\lim s_n = +\infty\) and \(k < 0\), then \(\lim(k s_n) = -\infty\).
Proof. Note \(-k > 0\). Thus \(\lim(-k s_n) = +\infty\) from part (a). Now using part (b) we can conclude \(\lim\big(-(-k s_n)\big) = \lim(k s_n) = -\infty\). \(\square\)
Exercise 9.11.
Exercise 9.11. (a) Show that if \(\lim s_n = +\infty\) and \(\inf\{t_n : n \in \mathbb{N}\} > -\infty\), then \(\lim(s_n + t_n) = +\infty\).
Proof. Let \(M > 0\). Since \(\inf\{t_n : n \in \mathbb{N}\} > -\infty\), the sequence \((t_n)\) is bounded below, thus there exists \(m\) such that \(m \le t_n\) for all \(n \in \mathbb{N}\). Since \(\lim s_n = +\infty\), there exists \(N \in \mathbb{N}\) such that for all \(n > N\), \(s_n > M - m\), where \((M - m) > 0\). Then \[ s_n + t_n > (M - m) + m = M \;\Longrightarrow\; \lim(s_n + t_n) = +\infty. \qquad \square \]
Exercise 9.11. (b) Show that if \(\lim s_n = +\infty\) and \(\lim t_n > -\infty\), then \(\lim(s_n + t_n) = +\infty\).
Proof.
Case \(\lim t_n = +\infty\). Let \(M > 0\). Let \(N_1 \in \mathbb{N}\) and \(N_2 \in \mathbb{N}\) such that \[ \forall n > N_1 \Rightarrow s_n > \tfrac{M}{2}, \qquad \forall n > N_2 \Rightarrow t_n > \tfrac{M}{2}. \] Let \(N = \max(N_1, N_2)\). Then for all \(n > N\), \(s_n + t_n > \frac{M}{2} + \frac{M}{2} = M\), so \(\lim(s_n + t_n) = +\infty\).
Case \(-\infty < \lim t_n < +\infty\). Let \(M > 0\). Since \(\lim t_n > -\infty\), there exists \(N_1 \in \mathbb{N}\) such that for all \(n > N_1\), \(|t_n - \lim t_n| < \frac{M}{2}\), hence \(t_n > \lim t_n - \frac{M}{2}\) for all \(n > N_1\). Since \(\lim s_n = +\infty\), there exists \(N_2\) such that for all \(n > N_2\), \(s_n > \frac{3M}{2} - \lim t_n\). Let \(N = \max\{N_1, N_2\}\). Then for all \(n > N\), \[ s_n + t_n > \left(\tfrac{3M}{2} - \lim t_n\right) + \left(\lim t_n - \tfrac{M}{2}\right) = M \;\Longrightarrow\; \lim(s_n + t_n) = +\infty. \qquad \square \]
Exercise 9.11. (c) Show that if \(\lim s_n = +\infty\) and if \((t_n)\) is a bounded sequence, then \(\lim(s_n + t_n) = +\infty\).
Proof. Since \((t_n)\) is bounded, there exist \(m_1, m_2\) such that \(m_1 \le t_n \le m_2\) for all \(n \in \mathbb{N}\). Let \(M > 0\). Since \(\lim s_n = +\infty\), there exists \(N \in \mathbb{N}\) such that for all \(n > N\), \(s_n > |m_1| + M > 0\). Note \(t_n \ge m_1\) for all \(n\) and \(s_n > |m_1| + M\) for all \(n > N\). Thus \[ s_n + t_n > |m_1| + M + m_1 \ge M \;\Longrightarrow\; \lim(s_n + t_n) = +\infty, \] since \(|m_1| + m_1 \ge 0\). \(\square\)
Exercise 9.12. Assume all \(s_n \ne 0\) and that the limit \(L = \lim\left|\dfrac{s_{n+1}}{s_n}\right|\) exists.
Exercise 9.12. (a) Show that if \(L < 1\), then \(\lim s_n = 0\).
Proof. Note \(0 \le |s_n|\) for all \(n\). Let \(a = \frac{L + 1}{2}\), so that \(0 \le L < a < 1\), and let \(\varepsilon = a - L > 0\). There exists \(N\) such that for all \(n \ge N\), \[ \left|\,\left|\tfrac{s_{n+1}}{s_n}\right| - L\,\right| < \varepsilon = a - L \;\Longrightarrow\; \left|\tfrac{s_{n+1}}{s_n}\right| < a \;\Longrightarrow\; |s_{n+1}| < a\,|s_n|. \] Then \[ |s_{N+1}| < a\,|s_N|, \qquad |s_{N+2}| < a\,|s_{N+1}| < a^2\,|s_N|, \qquad \dots \] so by induction \(|s_n| < a^{\,n - N}\,|s_N|\) for all \(n > N\). Since \(0 < a < 1\), \(a^{\,n-N} \to 0\), so \[ 0 \le |s_n| < a^{\,n - N}\,|s_N|, \] and by the Squeeze Lemma \(\lim |s_n| = 0\). This implies \(\lim s_n = 0\). \(\square\)
Exercise 9.12. (b) Show that if \(L > 1\), then \(\lim |s_n| = +\infty\).
Proof. Let \(t_n = \frac{1}{|s_n|}\). Note \(s_n \ne 0\) for all \(n\), so \(t_n\) is well-defined. Then \[ \left|\frac{t_{n+1}}{t_n}\right| = \frac{1/|s_{n+1}|}{1/|s_n|} = \frac{|s_n|}{|s_{n+1}|} = \frac{1}{\left|s_{n+1}/s_n\right|}, \] so \[ \lim \left|\frac{t_{n+1}}{t_n}\right| = \frac{1}{L}. \] Since \(L > 1\), we have \(\frac{1}{L} < 1\), so \(\lim t_n = 0\) from what we proved in part (a). Since \(\lim t_n = 0\) and \(t_n = \frac{1}{|s_n|}\), it follows that \(\lim |s_n| = +\infty\). \(\square\)