Exercises §10 — Ross, Elementary Analysis, 2nd Edition
Exercise 10.2. Prove that all bounded monotone decreasing sequences converge.
Proof. Let \((s_n)\) be a bounded decreasing sequence, let \(S = \{s_n : n \in \mathbb{N}\}\), and let \(\alpha = \inf S\). Since \((s_n)\) is bounded, \(S\) is a bounded nonempty subset of \(\mathbb{R}\), so \(\alpha = \inf S\) exists in \(\mathbb{R}\). We show \(\lim s_n = \alpha\).
Let \(\varepsilon > 0\). Since \(\alpha = \inf S\), the number \(\alpha + \varepsilon\) is not a lower bound of \(S\), so there exists \(N \in \mathbb{N}\) such that \(s_N < \alpha + \varepsilon\). Since \((s_n)\) is decreasing, \(s_n \le s_N\) for all \(n \ge N\); and since \(\alpha\) is a lower bound of \(S\), \(\alpha \le s_n\) for all \(n\). Thus for all \(n \ge N\), \[ \alpha \le s_n \le s_N < \alpha + \varepsilon \;\Longrightarrow\; |s_n - \alpha| < \varepsilon. \] Therefore \(\lim s_n = \alpha\). \(\square\)
Exercise 10.5. Prove that if \((s_n)\) is an unbounded decreasing sequence, then \(\lim s_n = -\infty\).
Proof. Let \(M < 0\), and let \(S = \{s_n : n \in \mathbb{N}\}\). Since \((s_n)\) is decreasing, \(s_n \le s_1\) for all \(n\), so \(S\) is bounded above by \(s_1\). Since \(S\) is unbounded, it must therefore be unbounded below. Hence there exists \(N \in \mathbb{N}\) such that \(s_N < M\). Since \((s_n)\) is decreasing, \(s_n \le s_N < M\) for all \(n \ge N\). Therefore \(\lim s_n = -\infty\). \(\square\)
Exercise 10.6. Let \((s_n)\) be a sequence such that \[ |s_{n+1} - s_n| < 2^{-n} \quad \text{for all } n \in \mathbb{N}. \]
Exercise 10.6. (a) Prove that \((s_n)\) is a Cauchy sequence and hence a convergent sequence.
Proof. From a previous exercise, \(\lim a^{\,n} = 0\) when \(0 < a < 1\); in particular \(\lim 2^{-n} = 0\). Assume \(m > n\) with \(m, n \in \mathbb{N}\). Telescoping and applying the triangle inequality with the hypothesis, \[ |s_m - s_n| = \left| \sum_{k=n}^{m-1} (s_{k+1} - s_k) \right| \le \sum_{k=n}^{m-1} |s_{k+1} - s_k| < \sum_{k=n}^{m-1} 2^{-k}. \] Since every term is positive, the partial sum is bounded by the full geometric tail: \[ \sum_{k=n}^{m-1} 2^{-k} < \sum_{k=n}^{\infty} 2^{-k} = 2^{-n} \sum_{j=0}^{\infty} 2^{-j} = 2^{-n} \cdot 2 = 2^{-n+1}. \] Thus \(|s_m - s_n| < 2^{-n+1}\) for all \(m > n\).
Now let \(\varepsilon > 0\). Since \(\lim 2^{-n+1} = 0\), there exists \(N \in \mathbb{N}\) such that \(2^{-N+1} < \varepsilon\). Then for all \(m > n \ge N\), \[ |s_m - s_n| < 2^{-n+1} \le 2^{-N+1} < \varepsilon. \] By symmetry the same bound holds when \(m < n\), and the case \(m = n\) is trivial. Hence \((s_n)\) is a Cauchy sequence, and by the completeness of \(\mathbb{R}\) it converges. \(\square\)
Exercise 10.6. (b) Is the result in (a) still true if we only assume \(|s_{n+1} - s_n| < \dfrac{1}{n}\) for all \(n \in \mathbb{N}\)?
Answer. No. Let \(s_n = \displaystyle\sum_{i=1}^{n} \frac{1}{i}\) be the \(n\)th partial sum of the harmonic series. Then \[ |s_{n+1} - s_n| = \frac{1}{n+1} < \frac{1}{n} \quad \text{for all } n \in \mathbb{N}, \] so the hypothesis holds. But the harmonic series diverges, so \(\lim s_n = +\infty\) and \((s_n)\) does not converge. The geometric bound \(2^{-n}\) in part (a) has a convergent tail sum, which is exactly what forces the sequence to be Cauchy; the harmonic bound \(\frac{1}{n}\) has a divergent tail, so the argument fails. \(\square\)
Exercise 10.7. Let \(S\) be a bounded nonempty subset of \(\mathbb{R}\) such that \(\sup S\) is not in \(S\). Prove that there is a sequence \((s_n)\) of points in \(S\) such that \(\lim s_n = \sup S\).
Proof. For each \(n \in \mathbb{N}\), since \(\sup S - \frac{1}{n} < \sup S\), the number \(\sup S - \frac{1}{n}\) is not an upper bound of \(S\). Hence the set \(\{x \in S : x > \sup S - \frac{1}{n}\}\) is nonempty, and we may choose \(s_n \in S\) with \(\sup S - \frac{1}{n} < s_n\). Since \(\sup S\) is an upper bound of \(S\) and \(s_n \in S\), we also have \(s_n \le \sup S\). Therefore \[ \sup S - \frac{1}{n} < s_n \le \sup S \quad \text{for all } n \in \mathbb{N}. \] Now \[ \lim\left( \sup S - \frac{1}{n} \right) = \sup S - 0 = \sup S, \qquad \lim \sup S = \sup S, \] so by the Squeeze Theorem \(\lim s_n = \sup S\). \(\square\)
Exercise 10.8. Let \((s_n)\) be an increasing sequence of positive numbers and define \(\sigma_n = \frac{1}{n}(s_1 + s_2 + \cdots + s_n)\). Prove that \((\sigma_n)\) is an increasing sequence.
Proof. Since \((s_n)\) is increasing, \(s_1 \le s_2 \le \cdots \le s_n \le s_{n+1}\) for all \(n \in \mathbb{N}\). In particular \(s_i \le s_{n+1}\) for every \(i \le n\), so \[ n\,\sigma_n = \sum_{i=1}^{n} s_i \le n\, s_{n+1} \;\Longrightarrow\; \sigma_n \le s_{n+1}. \] Now \[ \sigma_{n+1} = \frac{1}{n+1} \sum_{i=1}^{n+1} s_i = \frac{1}{n+1}\left( \sum_{i=1}^{n} s_i + s_{n+1} \right) = \frac{1}{n+1}\big( n\,\sigma_n + s_{n+1} \big). \] Using \(s_{n+1} \ge \sigma_n\), \[ \sigma_{n+1} \ge \frac{1}{n+1}\big( n\,\sigma_n + \sigma_n \big) = \frac{(n+1)\,\sigma_n}{n+1} = \sigma_n. \] Thus \(\sigma_{n+1} \ge \sigma_n\) for all \(n \in \mathbb{N}\), so \((\sigma_n)\) is increasing. \(\square\)