Solution Manual

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…