Exercises §11 — Ross, Elementary Analysis, 2nd Edition
Exercise 11.1. Let \(a_n = 3 + 2(-1)^n\) for all \(n \in \mathbb{N}\).
Exercise 11.1. (a) List the first eight terms of the sequence \((a_n)\).
Answer. Since \((-1)^n = -1\) for odd \(n\) and \((-1)^n = 1\) for even \(n\), we have \(a_n = 1\) for odd \(n\) and \(a_n = 5\) for even \(n\): \[ a_1 = 1, \quad a_2 = 5, \quad a_3 = 1, \quad a_4 = 5, \quad a_5 = 1, \quad a_6 = 5, \quad a_7 = 1, \quad a_8 = 5. \]
Exercise 11.1. (b) Give a subsequence that is constant (has only one value), and specify the selection function \(\sigma\).
Answer. Take the selection function \(\sigma(k) = 2k\) for \(k \in \mathbb{N}\), which is strictly increasing. The resulting subsequence is \[ a_{\sigma(k)} = a_{2k} = 3 + 2(-1)^{2k} = 3 + 2 = 5 \quad \text{for all } k \in \mathbb{N}, \] so \((a_{2k})\) is the constant subsequence with value \(5\). \(\square\)
Exercise 11.8. Prove that \(\liminf s_n = -\limsup(-s_n)\) for every sequence \((s_n)\).
Proof. Recall Definition 10.6: for a sequence \((s_n)\), \[ \limsup s_n = \lim_{N \to \infty} \sup\{s_n : n > N\}, \qquad \liminf s_n = \lim_{N \to \infty} \inf\{s_n : n > N\}. \]
We also use the result of Exercise 5.4: if \(S\) is a nonempty subset of \(\mathbb{R}\) and \(-S = \{-s : s \in S\}\), then \[ \inf S = -\sup(-S). \]
For each \(N \in \mathbb{N}\) the set \(\{s_n : n > N\}\) is nonempty, so applying the result of Exercise 5.4 with \(S = \{s_n : n > N\}\), and noting \(-S = \{-s_n : n > N\}\), gives \[ \inf\{s_n : n > N\} = -\sup\{-s_n : n > N\}. \] Therefore \[ \begin{aligned} \liminf s_n &= \lim_{N \to \infty} \inf\{s_n : n > N\} && \text{[Def.\ 10.6]} \\ &= \lim_{N \to \infty} \big( -\sup\{-s_n : n > N\} \big) && \text{[Exercise 5.4]} \\ &= -\lim_{N \to \infty} \sup\{-s_n : n > N\} && \text{[limit of a negation]} \\ &= -\limsup(-s_n). && \text{[Def.\ 10.6]} \end{aligned} \] \(\square\)