Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 8

Problem

If $\sum a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges.

Answer

Since $b_n$ is monotonic and bounded, $b_n \rightarrow b$. If $b_n$ is increasing, then $\sum a_n (b-b_n)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b - \sum a_n (b - b_n)$ also converges.

If $b_n$ is decreasing, then $\sum a_n (b_n-b)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b + \sum a_n (b_n - b)$ also converges. 

Comments

Post a Comment

Popular posts from this blog

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

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 4

Problem Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p)=f(p)$ for all $p \in E$, prove that $g(p)=f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Answer For $y \in f(X)$, there exists $x \in X$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $E$ is dense in $X$, there exists $z \in E$ such that $z \in N_\delta(x)$, which implies $f(x) \in N_\epsilon(f(z))$. This means $f(E)$ is dense in $f(X)$. For $p \in X$ and $\epsilon >0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2$. For $q \in E$ satisfying $q \in N_\delta(p)$, $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q)...
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 14

Problem Give an example of an open cover of the segment $(0,1)$ which has no finite subcover. Answer Consider a family of sets $\{(1/n,1)\}_{n=2}^\infty$. For any $x \in (0,1)$, there exists $N \in \mathbb{N}$ such that $1/N < x$, which means $x \in (1/N,1)$. This means that $\{(1/n,1)\}_{n=2}^\infty$ covers $(0,1)$. Consider an arbitrary finite subcover $\{(1/n_1,1),\cdots,(1/n_K)\}$ of $\{(1/n,1)\}_{n=2}^\infty$. Then $\bigcup_{k=1}^K(1/n_k,1)=(1/n_\ast,1)$ where $n_\ast=\max(n_1,\cdots,n_K)$, which cannot cover $(0,1)$.
Read more