Category: Understanding Analysis

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.7 0

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.7

Exercise 2.7.2 (a) Converges. Since $\sum\limits_{n=0}^\infty \dfrac{1}{2^n}$ converges, see Example 2.7.5 (Geometric Series), and $\dfrac{1}{2^n+n}\leqslant \dfrac{1}{2^n}$, by Theorem 2.7.4 (Comparison Test), we have $\sum\limits_{n=0}^\infty \dfrac{1}{2^n+n}$ converges. (b) Converges. Note that $\sum\limits_{n=0}^\infty \dfrac{1}{n^2}$ converges and...

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.6 0

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.6

Exercise 2.6.1 Let $\varepsilon > 0$ be arbitrary. Since $(x_n)$ converges to $x$, there exits $N\in\mathbf N$ such that \begin{equation}\label{eq2.6.1.1}|x_n-x|<\frac{\varepsilon}{2}\end{equation} for all $n\geqslant N$. Whenever $n,m\geqslant N$, we have\begin{align*}|x_n-x_m|\leqslant &~|x_n-x|+|x-x_m|\\ \text{by }\eqref{eq2.6.1.1}\quad< &~ \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\end{align*}Hence...

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.5 2

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.5

Exercise 2.5.1 (a) Impossible by Theorem 2.5.5. (b) Consider the sequence given by\[a_n=\frac{1+(-1)^n+\frac{1}{n}}{2}.\]Then we have\[a_{2n}=1+\frac{1}{4n},~a_{2n-1}=\frac{1}{2(2n-1)}.\]It is clear that $(a_{2n})$ converges to $1$ and $(a_{2n-1})$ converges to $0$. Moreover, it is easy to check that...

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.4 1

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.4

Exercise 2.4.1 (a) We show by induction that $(x_n)$ is decreasing by proving $x_n > x_{n+1}$. Note that $x_1=3$ and $x_2=\dfrac{1}{4-3}=1$. Hence $x_1>x_2$. Suppose now that\[3=x_1> x_2>\cdots >x_n> x_{n+1},\]we show that $x_{n+1}> x_{n+2}$. By...

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.3 1

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.3

Exercise 2.3.1 (a) Let $\varepsilon > 0$ be arbitrary. Since $(x_n)\to 0$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $$x_n=|x_n|<\varepsilon^2.$$Hence $\sqrt{x_n}<\varepsilon$. Therefore, for all $n>N$, we also have\[|\sqrt{x_n}-0|=\sqrt{x_n}<\varepsilon.\]Thus we have $(\sqrt{x_n})\to...

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.2 1

Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.2

Exercise 2.2.1 Example: Take the sequence (1,0,1,0,1,0,1,0,…….). To check this sequence verconges $0$, take e.g. $\varepsilon=2$. The previous example defines a divergent sequence. It is not hard to see the sequence in our example...