Author: Linearity

Exercise 2.7.1 See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.1 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…

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|\\…

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…

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…

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…

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 example defines a divergent sequence. It is not hard to see the sequence…

Exercise 1.2.1 See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.1 Exercise 1.2.2 We prove it by contradiction. Suppose there exists a rational number $r$ such that $2^r=3$. Let $r=p/q$, where…