Exercises §8 — Ross, Elementary Analysis, 2nd Edition
Exercise 8.4. Let \((t_n)\) be a bounded sequence, i.e. there exists \(M > 0\) such that \(|t_n| \le M\) for all \(n\), and let \((s_n)\) be a sequence such that \(\lim s_n = 0\). Prove \(\lim(s_n t_n) = 0\). (Do this proof without using the limit theorems of the next section.)
Proof. Let \(\varepsilon > 0\). We want to show \(|s_n t_n - 0| < \varepsilon\) for large enough \(n\), i.e. \(|s_n t_n| < \varepsilon\).
Since \(\lim s_n = 0\), choose \(N\) such that for all \(n > N\) we have \(|s_n - 0| < \frac{\varepsilon}{M}\). Then for all \(n > N\), \[ M\,|s_n - 0| < \varepsilon \quad\Longrightarrow\quad |s_n M - 0| < \varepsilon. \] Note that \(|t_n| \le M\) for all \(n\), so \(|s_n t_n| \le |s_n M|\) for all \(n\). Hence for all \(n > N\), \[ |s_n t_n| \le |s_n M| < \varepsilon, \] so \(\lim (s_n t_n) = 0\). \(\square\)
Exercise 8.5. (Squeeze Lemma)
Exercise 8.5. (a) Consider sequences \((a_n)\), \((b_n)\), and \((s_n)\) such that \(a_n \le s_n \le b_n\) for all \(n \in \mathbb{N}\) and \(\lim a_n = \lim b_n = s\). Prove \(\lim s_n = s\).
Proof. We want to show that for all \(\varepsilon > 0\), \(|s_n - s| < \varepsilon\) for large enough \(n\).
Let \(\varepsilon > 0\). Choose \(N \in \mathbb{N}\) such that \(|a_n - s| < \varepsilon\) and \(|b_n - s| < \varepsilon\) for all \(n > N\). Note that \[ a_n - s \le s_n - s \le b_n - s. \] From \(|a_n - s| < \varepsilon\) we get \(-\varepsilon < a_n - s\), and from \(|b_n - s| < \varepsilon\) we get \(b_n - s < \varepsilon\). Therefore, for all \(n > N\), \[ -\varepsilon < a_n - s \le s_n - s \le b_n - s < \varepsilon, \] so \(-\varepsilon < s_n - s < \varepsilon\), i.e. \(|s_n - s| < \varepsilon\). Hence \(\lim s_n = s\). \(\square\)
Exercise 8.5. (b) Suppose \((s_n)\) and \((t_n)\) are sequences such that \(|s_n| \le t_n\) for all \(n\) and \(\lim t_n = 0\). Prove \(\lim s_n = 0\).
Proof. Note that \(|s_n| \le t_n\) for all \(n\) implies \(t_n \ge 0\) for all \(n\), since \(|s_n| \ge 0\) for all \(n\). Thus \(-t_n \le s_n \le t_n\) for all \(n\).
Let \(a_n = -t_n\) and \(b_n = t_n\) for all \(n\). Then \(a_n \le s_n \le b_n\) for all \(n\), and \(\lim a_n = -\lim t_n = 0\) and \(\lim b_n = \lim t_n = 0\). By the Squeeze Lemma (part (a)), \(\lim s_n = 0\). \(\square\)
Exercise 8.6. Let \((s_n)\) be a sequence in \(\mathbb{R}\). Prove \(\lim s_n = 0\) if and only if \(\lim |s_n| = 0\).
Proof.
(\(\Rightarrow\)) Suppose \(\lim s_n = 0\). Let \(\varepsilon > 0\) and choose \(N\) such that \(|s_n - 0| < \varepsilon\) for all \(n > N\). Note that \(|\,|s_n| - 0\,| = |s_n| = |s_n - 0|\) for all \(n\), so \(|\,|s_n| - 0\,| < \varepsilon\) for all \(n > N\). Hence \(\lim |s_n| = 0\).
(\(\Leftarrow\)) Suppose \(\lim |s_n| = 0\). Let \(\varepsilon > 0\) and choose \(N\) such that \(|\,|s_n| - 0\,| < \varepsilon\) for all \(n > N\). Note that \(|\,|s_n| - 0\,| = |s_n| = |s_n - 0|\). Thus \(|s_n - 0| < \varepsilon\) for all \(n > N\), so \(\lim s_n = 0\). \(\square\)
Exercise 8.9. Let \((s_n)\) be a sequence that converges, and let \(a \in \mathbb{R}\). Write \(s = \lim s_n\).
Exercise 8.9. (a) Show that if \(s_n \ge a\) for all but finitely many \(n\), then \(\lim s_n \ge a\).
Proof. Assume for contradiction that \(\lim s_n < a\). Then \(a - s > 0\), where \(s = \lim s_n\). Let \[ \varepsilon = \frac{a - s}{2} > 0. \] There exists \(N_1\) such that for all \(n > N_1\), \(|s_n - s| < \frac{a - s}{2}\). Then \[ -\frac{a - s}{2} < s_n - s < \frac{a - s}{2}. \] Adding \(s\) throughout, \[ \frac{3s - a}{2} < s_n < \frac{s + a}{2}. \] Since \(\lim s_n < a\), we have \(s < a\), so \(\frac{s + a}{2} < a\). Hence \(s_n < a\) for all \(n > N_1\), which contradicts \(s_n \ge a\) for all but finitely many \(n\). Therefore \(\lim s_n \ge a\). \(\square\)
Exercise 8.9. (b) Show that if \(s_n \le b\) for all but finitely many \(n\), then \(\lim s_n \le b\).
Proof. Note that \(-s_n \ge -b\) for all but finitely many \(n\), so by part (a), \(\lim(-s_n) \ge -b\). Since \(\lim(-s_n) = -\lim s_n\), we get \(-\lim s_n \ge -b\), hence \(\lim s_n \le b\). \(\square\)
Exercise 8.9. (c) Show that if all but finitely many \(s_n\) belong to \([a, b]\), then \(\lim s_n\) belongs to \([a, b]\).
Proof. Since \(s_n \ge a\) for all but finitely many \(n\), part (a) gives \(\lim s_n \ge a\). Since \(s_n \le b\) for all but finitely many \(n\), part (b) gives \(\lim s_n \le b\). Therefore \(a \le \lim s_n \le b\), i.e. \(\lim s_n \in [a, b]\). \(\square\)
Exercise 8.10. Let \((s_n)\) be a convergent sequence and suppose \(\lim s_n > a\). Prove that there exists \(N\) such that \(n > N\) implies \(s_n > a\). Write \(s = \lim s_n\).
Proof. Let \(\varepsilon = s - a > 0\). There exists \(N\) such that for all \(n > N\), \(|s_n - s| < s - a\). Then \[ -(s - a) < s_n - s < s - a. \] Adding \(s\) throughout, \[ a < s_n < 2s - a. \] In particular, \(s_n > a\) for all \(n > N\). \(\square\)