Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 3

Problem

If $s_1 = \sqrt{2}$, and

$$ s_{n+1}=\sqrt{2 + \sqrt{s_n}}\;\;\;(n=1,2,3,\cdots),$$

prove that $\{s_n\}$ converges, and that $s_n < 2$ for $n=1,2,3,\cdots$.

Answer

Note that $s_1 < 2$. If $s_n < 2$ then

$$ s_{n+1} = \sqrt{2 + \sqrt{s_n}} < \sqrt{2+\sqrt{2}} < 2. $$

By mathematical induction, $s_n < 2$ for $n=1,2,3,\cdots$.

Now we claim that $s_n$ is increasing, and hence convergent.

Note that $s_1 = \sqrt{2} < \sqrt{2 + \sqrt{2}}$. If $s_{n-1} < s_n$ then

$$ s_{n+1} = \sqrt{2+\sqrt{s_n}} > \sqrt{2 + \sqrt{s_{n-1}}} = s_n$$.

By mathematical induction, $s_n$ is increasing.

Comments

Post a Comment

Popular posts from this blog

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 29

Problem Prove that every open set in $\mathbb{R}^1$ is the union of an at most countable collection of disjoint segments. Answer Let $K_l:=\{[m,m+1/2^l)\;|\; m \in \mathbb{Z}/2^l\}$. Then $K_l$ is a collection of coutable disjoint segments with length $1/2^l$. For open set $G \subset \mathbb{R}^1$, define $$S_0:=\{Q \in K_0\;|\; Q \subset G\}.$$ $$S_n:=\{Q \in K_n\;|\; Q \subset G, \text{$Q \not\subset Q'$ for some $Q' \in \bigcup_{i=1}^{n-1}K_i$}  \}.$$ $$S:=\bigcup_{n=1}^\infty S_n.$$ Then $S$ is at most countable collection of disjoint segments and $\bigcup_{Q \in S}Q \subset G$. For $x \in G$, there exists $N \in \mathbb{N}$ such that $N_{1/2^N}(x) \subset G$. Then, we can find $m \in \mathbb{Z}/2^N$ such that $m \leq x$ and $m \in N_{1/2^N}(x)$. This means $x \in Q$ for some $Q \in \bigcup_{n=1}^N S_n$. Hence $G \subset \bigcup_{Q \in S}Q$.
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 16

Problem Regard $\mathbb{Q}$, the set of all rational numbers, as a metric space, with $d(p,q)=|p-q|$. Let $E$ be the set of all $p \in \mathbb{Q}$ such that $2 < p^2 < 3$. Show that $E$ is closed and bounded in $\mathbb{Q}$, but that $E$ is not compact. Is $E$ open in $\mathbb{Q}$? Answer (1) Closed: $E=\mathbb{Q} \cap ((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3})) = \mathbb{Q} \cap ([-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}])$. Define $F:=[-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}]$  $\Rightarrow$  $F^c$ is open in $\mathbb{R}$  $\Rightarrow$  $\mathbb{Q} \cap F^c$ is open in $\mathbb{Q}$ by Theorem 2.30  $\Rightarrow$  $\mathbb{Q} \cap F$ is closed in $\mathbb{Q}$. (2) Bounded: $d(p,0) < \sqrt{3}$ for all $p \in E$, which means $E$ is bounded in $\mathbb{Q}$. (3) Not compact: $E$ is not compact in $\mathbb{R}$, and hence is not compact in $\mathbb{Q}$ by Theorem 2.33. (4) Open: Since $((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3}))$ is ...
Read more

Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 1

Problem Discuss whether the following sets are open or closed: (a) (1,2) in $\mathbb{R}$ (b) [2,3] in $\mathbb{R}$ (c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$ (d) $\mathbb{R}^n$ in $\mathbb{R}^n$ (e) A hyperplane in $\mathbb{R}^n$ (f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$ (g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$ (h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$. Answer (a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$. (b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$. (c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$. To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means ...
Read more