Sem 2 - Sequences, monotony, boundedness, limits

Exercise 1.

Prove using the $\epsilon$-definition that $\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}$ $\forall \epsilon >0,\exists N_{\epsilon }\in \mathbb{N}\text{ s.t. }|\frac{1}{\sqrt{n}}|<\epsilon ,\forall n\ge N_{\epsilon }$ Let $\epsilon >0$. Take $N_{\epsilon }=[\frac{1}{{\epsilon }^2}]+1$ $\frac{1}{\sqrt{n}}<\epsilon ,n\ge N_{\epsilon }$ $\frac{1}{n}<{\epsilon }^2,n>\frac{1}{{\epsilon }^2},\forall n\ge N_{\epsilon }$ $n\ge N_{\epsilon }>\frac{1}{{\epsilon }^2}$

Exercise 2.

A. $x_n=\sqrt{n+1}-\sqrt{n}$

$x_n=\sqrt{n+1}-\sqrt{n}=+\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ $x_n>0,x_n<x_0=1,n\ge 0⟹x_n\in [0,1](\ast )$ $\frac{x_n+1}{x_n}=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}<1,\sqrt{n}<\sqrt{n+2}$ $x_{n+1}<x_n$ (decreasing) $(\ast \ast )$ $(\ast )+(\ast \ast )⟹(x_n)convergent,x_n\to 0$

B. $x_n=\frac{1}{1\times 2}+\dots +\frac{1}{n(n+1)}$

$\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\dots +\frac{(n+1)-n}{n(n+1)}=$ $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\dots +\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}<1⟹$ $⟹x_n\in (0,1),\text{ bounded}(\ast )$ $x_{n+1}-x_n=\frac{1}{(n+1)(n+2)}>0,x_{n+1}>x_n(\ast \ast )$ $(\ast ),(\ast \ast )⟹\text{ convergent}$

C. $x_n=\frac{2^n}{n!}$

$x_n>0$ $\frac{x_{n+1}}{x_n}=\frac{2^{n+1}}{(n+1)!}\times \frac{n!}{2^n}=\frac{2}{n+1}<1$ $x_n\in (0,x_1],x_{n+1}<x_n,n\ge 2⟹(x_n)\text{ convergent}$

Exercise 3.

A. $\sqrt{n}(\sqrt{n+1}-\sqrt{n})$

$\underset{n\to \infty }{\lim }\sqrt{n}(\sqrt{n+1}-\sqrt{n})\stackrel{\infty \times 0}{=}\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}=\frac{1}{2}$

B. $\frac{2^n+(-1{)}^n}{3^n}$

$\underset{n\to \infty }{\lim }\frac{2^n+(-1{)}^n}{3^n}=\underset{n\to \infty }{\lim }{\left(\frac{2}{3}\right)}^n+\underset{n\to \infty }{\lim }(-\frac{1}{3}{)}^n=0+0=0$

C. $\sqrt[n]{n}$

$\underset{n\to \infty }{\lim }\sqrt[n]{n}=\underset{n\to \infty }{\lim }{n}^{\frac{1}{n}}\stackrel{{\infty }^0}{=}$

Using $e$ for ${\infty }^0$ $e^{\ln x}=x,f^g=e^{ln{f}^g}=e^{g\ln f}$

$n^{\frac{1}{n}}=e^{\frac{\ln n}{n}}\to e^0=1$ $\underset{n\to \infty }{\lim }\frac{\ln n}{n}\stackrel{S.C.}{=}\underset{n\to \infty }{\lim }\frac{\ln \frac{n+1}{n}}{n+1-n}\dots$

E. $(2^n+3^n{)}^{\frac{1}{n}}={\left(3^n\left(1+{\left(\frac{2}{3}\right)}^n\right)\right)}^{\frac{1}{n}}=3\times 1$

$(a_1^n+a_2^n+\dots +a_k^n{)}^{\frac{1}{n}}$ $a_i=max\{a_1,\dots ,a_k\}$ $=a_i{\left[{\left(\frac{a_1}{a_i}\right)}^n+\dots +{\left(\frac{a_k}{a_i}\right)}^n\right]}^{\frac{1}{n}}$ $(\frac{a_j}{a_i}{)}^n\to 0,j\ne i$

Exercise 5.

A. $(\frac{2n+1}{2n-1}{)}^n$

$=[(1+\frac{2}{2n-1}{)}^{\frac{2n-1}{2}}{]}^{\frac{2n}{2n-1}}$

B. $n(\ln (n+2)-\ln (n+1)$

$n\ln \frac{n+2}{n+1}=\ln (\frac{n+2}{n+1}{)}^n=\ln \underset{\to 3}{[(1+\frac{1}{n+1}{)}^{n+1}{]}^{\frac{n}{n+1}\to 1}}\to \ln e=1$

Exercise 6.

A.

$\underset{n\to \infty }{\lim }\frac{x_1+\dots +x_n}{n}\stackrel{S.C.}{=}\underset{n\to \infty }{\lim }\frac{x_{n+1}}{n+1-n}=\underset{n\to \infty }{\lim }{x}_n$ $\exists \underset{n\to \infty }{\lim }\frac{x_1+\dots +x_n}{b},\text{ but}\nexists \underset{n\to \infty }{\lim }{x}_n$ $x_n=(-1{)}^n,\nexists \underset{n\to \infty }{\lim }{x}_n$

$$\frac{x_1+x_2+\dots +x_n}{n}=\{\begin{array}{l}0,\text{n even}\\ -\frac{1}{n},\text{ n odd}\end{array}⟹\lim \frac{x_1+\dots x_n}{n}=0$$

B.

$\underset{n\to \infty }{\lim }\frac{1+\frac{1}{2}+\dots +\frac{1}{n}}{n}\stackrel{S.C.}{=}\underset{n\to \infty }{\lim }\frac{\frac{1}{n+1}}{1}=0$ $\underset{n\to \infty }{\lim }\frac{1+\frac{1}{2}+\dots +\frac{1}{n}}{\ln n}\stackrel{S.C.}{=}\underset{n\to \infty }{\lim }\frac{\frac{1}{n+1}}{\ln \frac{n+1}{n}}=\underset{n\to \infty }{\lim }\frac{}{n\ln \frac{n+1}{n}}\times \frac{n}{n+1}=\lim \frac{1}{\ln {\left(1+\frac{1}{n}\right)}^n}\times \frac{n}{n+1}=1\times 1=1$

Exercise 7.

$l=\underset{n\to \infty }{\lim }\frac{x_{n+1}}{x_n}$ $\sqrt[n]{x_n}=x_n^{\frac{1}{n}}=e^{\frac{1}{n}\ln x_n}=e^{\frac{\ln x_n}{n}}\to e^{\ln e}=e$ $\underset{n\to \infty }{\lim }\frac{\ln x_n}{n}\stackrel{S.C.}{=}\underset{n\to \infty }{\lim }\frac{\ln x_{n+1}-\ln x_n}{n+1-n}=\underset{n\to \infty }{\lim }\ln \frac{x_{n+1}}{x_n}=\ln \underset{n\to \infty }{\lim }\frac{x_{n+1}}{x_n}=\ln e$

Exercise 8.

A.

$\frac{n}{\sqrt[n]{n!}}=\sqrt[n]{\frac{n^n}{n!}}=\sqrt[n]{x_n}⟹\underset{n\to \infty }{\lim }\sqrt[n]{x_n}=\underset{n\to \infty }{\lim }\frac{x_{n+1}}{x_n}=e$ $\underset{n\to \infty }{\lim }\frac{x_{n+1}}{x_n}=\underset{n\to \infty }{\lim }\frac{(n+1{)}^{n+1}}{(n+1)!}\times \frac{n!}{n^n}$ $\underset{n\to \infty }{\lim }\frac{(n+1{)}^n}{n^n}=\underset{n\to \infty }{\lim }{\left(\frac{n+1}{n}\right)}^n=\underset{n\to \infty }{\lim }(1+\frac{1}{n}{)}^n=e$