Exercises §12 — Ross, Elementary Analysis, 2nd Edition

Exercises

Exercise 12.1 · Exercise 12.10 · Exercise 12.11

Exercise 12.1. Let $(s_n)$ and $(t_n)$ be sequences and suppose there exists $N_0$ such that $s_n \le t_n$ for all $n > N_0$. Show $\liminf s_n \le \liminf t_n$ and $\limsup s_n \le \limsup t_n$.

Proof. Recall Definition 10.6: for a sequence $(x_n)$, \[ \limsup x_n = \lim_{N \to \infty} \sup\{x_n : n > N\}, \qquad \liminf x_n = \lim_{N \to \infty} \inf\{x_n : n > N\}. \]

For every $N \ge N_0$ we have $\{n : n > N\} \subseteq \{n : n > N_0\}$, so $s_n \le t_n$ holds for every $n > N$. Taking suprema and infima over the tail $\{n : n > N\}$ preserves the inequality: \[ \sup\{s_n : n > N\} \le \sup\{t_n : n > N\}, \qquad \inf\{s_n : n > N\} \le \inf\{t_n : n > N\}. \]

Passing to the limit as $N \to \infty$ (limits preserve $\le$) and applying Definition 10.6 gives \[ \begin{aligned} \limsup s_n &= \lim_{N \to \infty} \sup\{s_n : n > N\} && \text{[Def.\ 10.6]} \\ &\le \lim_{N \to \infty} \sup\{t_n : n > N\} && \text{[$\sup$ inequality on the tail, limits preserve $\le$]} \\ &= \limsup t_n, && \text{[Def.\ 10.6]} \end{aligned} \] and likewise \[ \begin{aligned} \liminf s_n &= \lim_{N \to \infty} \inf\{s_n : n > N\} && \text{[Def.\ 10.6]} \\ &\le \lim_{N \to \infty} \inf\{t_n : n > N\} && \text{[$\inf$ inequality on the tail, limits preserve $\le$]} \\ &= \liminf t_n. && \text{[Def.\ 10.6]} \end{aligned} \] $\square$

Exercise 12.10. Prove $(s_n)$ is bounded if and only if $\limsup |s_n| < +\infty$.

Proof. ($\Rightarrow$) Suppose $(s_n)$ is bounded. Then there exist $M_1, M_2 \in \mathbb{R}$ such that $M_1 \le s_n \le M_2$ for all $n \in \mathbb{N}$. Let $M = \max\{|M_1|, |M_2|\}$. Then $|s_n| \le M$ for all $n \in \mathbb{N}$, so $(|s_n|)$ is bounded and \[ \limsup |s_n| \le M < +\infty. \]

($\Leftarrow$) Suppose $\limsup |s_n| < +\infty$. Since $0 \le |s_n|$ for all $n \in \mathbb{N}$, we have $0 \le \limsup |s_n|$.

Case 1: $\limsup |s_n| = 0$. Then $0 \le \liminf |s_n| \le \limsup |s_n| = 0$, so $\liminf |s_n| = \limsup |s_n| = 0$, which gives $\lim |s_n| = 0$. A convergent sequence is bounded (Theorem 9.1), so $(|s_n|)$ is bounded, and hence so is $(s_n)$.

Case 2: $0 < \limsup |s_n| < +\infty$. Let $L = \limsup |s_n|$ and take $\varepsilon = 1$. By definition of $\limsup$ there exists $N \in \mathbb{N}$ such that \[ |s_n| < L + 1 \quad \text{for all } n \ge N. \] Thus only the finitely many terms $|s_1|, \ldots, |s_{N-1}|$ can exceed $L + 1$. Let \[ M = \max\{|s_1|, \ldots, |s_{N-1}|, \, L + 1\}. \] Then $0 \le |s_n| \le M$ for all $n$, so $(|s_n|)$ is bounded, and $-M \le s_n \le M$ for all $n$, which means $(s_n)$ is bounded. $\square$

Exercise 12.11. Let $(s_n)$ be any sequence of nonzero real numbers. Prove that \[ \liminf \left| \frac{s_{n+1}}{s_n} \right| \le \liminf |s_n|^{1/n}. \]

Proof. Let $L = \liminf \left| \dfrac{s_{n+1}}{s_n} \right|$ and $\alpha = \liminf |s_n|^{1/n}$. We must prove $\alpha \ge L$. This is immediate if $L = 0$ (since $\alpha \ge 0$), so assume $L > 0$. To prove $\alpha \ge L$ it suffices to prove $\alpha \ge L_1$ for any $L_1$ with $0 < L_1 < L$.

Since \[ L = \liminf \left| \frac{s_{n+1}}{s_n} \right| = \lim_{N \to \infty} \inf\left\{ \left| \frac{s_{n+1}}{s_n} \right| : n \ge N \right\} > L_1, \] there exists $N \in \mathbb{N}$ such that $\inf\left\{ \left| \dfrac{s_{n+1}}{s_n} \right| : n \ge N \right\} > L_1$. Thus \[ \left| \frac{s_{n+1}}{s_n} \right| > L_1 \quad \text{for all } n \ge N. \]

So for all $n > N$ we can write, telescoping the ratios, \[ |s_n| = \left| \frac{s_n}{s_{n-1}} \right| \cdot \left| \frac{s_{n-1}}{s_{n-2}} \right| \cdots \left| \frac{s_{N+1}}{s_N} \right| \cdot |s_N|. \] There are $n - (N+1) + 1 = n - N$ fractions, each exceeding $L_1$, so \[ |s_n| > L_1^{\,n-N} |s_N| \quad \text{for all } n > N. \] Let $a = L_1^{-N} |s_N| > 0$. Then $|s_n| \ge a\, L_1^{\,n}$, and therefore \[ |s_n|^{1/n} \ge a^{1/n} L_1. \] Since $\lim_{n \to \infty} a^{1/n} = 1$, taking $\liminf$ of both sides gives \[ \liminf |s_n|^{1/n} \ge L_1. \] As $L_1 \in (0, L)$ was arbitrary, we conclude \[ \liminf |s_n|^{1/n} \ge L = \liminf \left| \frac{s_{n+1}}{s_n} \right|. \qquad \square \]