Sequence. Boundedness and monotonicity of a sequence. Limit of a sequence.

A sequence of real numbers is defined as a function $f:\mathbb{N}\to \mathbb{R}$ defined on the set of all natural numbers. The number $f(n)$ is called the $n$-th term of the sequence $x_n$, and the formula $x_n=f(n)$ is called the general term formula of the sequence $(x_n{)}_{n\in \mathbb{N}}$.

A number $a$ is called the limit of the sequence $(x_n)n\in \mathbb{N}$ if for any $\epsilon >0$ there exists a number $N(\epsilon )$ such that for $n>N(\epsilon )$ the inequality $|xn+p-x_n|<\epsilon$ holds. In this case, the sequence itself is called convergent.

Cauchy Criterion.

For a sequence $(x_n)n\in \mathbb{N}$ to have a limit, it is necessary and sufficient that for any $\epsilon >0$ there exists a number $N(\epsilon )$ such that for $n>N(\epsilon )$ the inequality $|xn+p-x_n|<\epsilon$ holds for any $p\in \mathbb{N}$.

A sequence $(x_n)n\in \mathbb{N}$ is called infinitesimal if $limn\to \mathrm{\infty }=0$.

A sequence $(x_n)n\in \mathbb{N}$ is called infinitely large (converging to infinity), denoted as $limn\to \mathrm{\infty }=\mathrm{\infty }$, if for any number $E>0$ there exists a number $N(E)$ such that for $n>N(E)$ the inequality $|x_n|>E$ holds. If from some index onwards, all terms are positive (negative), then the next notation is used

$$\underset{n\to \mathrm{\infty }}{lim}=+\mathrm{\infty }\underset{n\to \mathrm{\infty }}{lim}=-\mathrm{\infty }$$

The number $a$ is called a limit point of the sequence $(x_n{)}_{n\in \mathbb{N}}$ if for any $\epsilon >0$ there exist infinitely many terms of this sequence satisfying the condition $|x_n-a|<\epsilon$.

Bolzano-Weierstrass Principle.

Any bounded sequence has at least one limit point.

The greatest (smallest) of the limit points of the sequence $(x_n{)}_{n\in \mathbb{N}}$ is called the upper (lower) limit of this sequence and is denoted by the symbol

$\underset{n\to \mathrm{\infty }}{\bar{\lim }}{x}_n$ ($\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}{x}_n$).

Examples.

In problems 1, 2, write the first five terms of the sequence:

1 $x_n=1+(-1{)}^n\frac{1}{n}.$

Solution.

$x_1=1+(-1{)}^1\frac{1}{1}=1-1=0;$

$x_2=1+(-1{)}^2\frac{1}{2}=1+\frac{1}{2}=1,5;$

$x_3=1+(-1{)}^3\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3};$

$x_4=1+(-1{)}^4\frac{1}{4}=1+\frac{1}{4}=1,25;$

$x_5=1+(-1{)}^5\frac{1}{5}=1-\frac{1}{5}=\frac{4}{5}=0,8.$

Answe: $0;1,5;\frac{2}{3};1,25;0,8.$

2. $x_n=\frac{3n+5}{2n-3}.$

Solution.

$x_1=\frac{3+5}{2-3}=-8;$

$x_2=\frac{6+5}{4-3}=11;$

$x_3=\frac{9+5}{6-3}=\frac{14}{3};$

$x_4=\frac{12+5}{8-3}=\frac{17}{5};$

$x_5=\frac{15+5}{10-3}=\frac{20}{7}.$

Answe: $-8;11;\frac{14}{3};\frac{17}{5};\frac{20}{7}.$

In problems 3, 4, write the general term formula of the sequence:

3. $-\frac{1}{2},,\frac{1}{3},,-\frac{1}{4},,\frac{1}{5},...$

Solution.

From the given sequence, we have:

$x_1=-\frac{1}{2};$

$x_2=\frac{1}{3};$

$x_3=-\frac{1}{4};$

$x_4=\frac{1}{5};$

Continuing this pattern, we find the general term of the sequence:

$$x_n=(-1{)}^n\frac{1}{n+1}.$$

Answe: $x_n=(-1{)}^n\frac{1}{n+1}.$

4. $0,2,0,2,...$

Solution.

From the given conditions, we have:

$x_1=0;$

$x_2=2;$

$x_3=0;$

$x_4=2;$

...

That is, for odd indices $x_{2k-1}=0,$

and for even indices $x_{2k}=2.$

The general term of the sequence can be expressed as $x_n=1+(-1{)}^n.$

Answer: $x_n=1+(-1{)}^n.$

1.220. $1,0,-3,0,5,0,-7,0,...$

Solution.

From the given conditions, we have:

$x_1=1;$

$x_2=0;$

$x_3=-3;$

$x_4=0;$

...

That is, for odd indices $x_{2k-1}=(2k-1)(-1{)}^{k+1};$ or $x_m=m(-1{)}^{\frac{m+1}{2}}=m\cos \frac{m-1}{2}\pi$

and for even indices $x_{2k}=0$ or $x_m=m\cos \frac{\pi (m-1)}{2}.$

The general term of the sequence can be expressed as $x_n=n\cos \frac{\pi (n-1)}{2}.$

Answer: $x_n=n\cos \frac{\pi (n-1)}{2}.$

In problems 5, 6, find the largest (smallest) term of the bounded sequence $(x_n{)}_{n\in \mathbb{N}}.$

5. $x_n=6n-n^2-5.$

Solution.

It is evident that $6n-n^2=n(6-n)<0$ for all $n\ge 6,$ thus $6n-n^2-5<-5.$ Let's write down several initial terms of the sequence:

$x_1=6-1-5=0;$

$x_2=12-4-5=3;$

$x_3=18-9-5=4;$

$x_4=24-16-5=3;$

$x_5=30-25-5=0;$

$x_6=36-36-5=-5;$

$x_7=42-49-5=-12$

...

The largest term of the sequence is $x_3=4.$

Answer: $4.$

6. $x_n=-\frac{n^2}{2^n}.$

Solution.

Let's write down several initial terms of the sequence:

$x_1=-\frac{1}{2};$

$x_2=-\frac{4}{4}=-1;$

$x_3=-\frac{9}{8};$

$x_4=-\frac{16}{16}=-1;$

$x_5=-\frac{25}{32};$

$x_6=-\frac{36}{64};$

......

We will show that for all $n\ge 3$, the inequality $|x_{n+1}|<|x_n|$ holds:

$$\frac{n^2}{2^n}>\frac{(n+1)}{2^{n+1}}=\frac{n^2+2n+1}{2^{n+1}}=\frac{n^2}{2\cdot 2^n}+\frac{n+1}{2^{n+1}}\Rightarrow$$

$$\Rightarrow \frac{n^2}{2^{n+1}}>\frac{n+1}{2^{n+1}}\Rightarrow n^2>n+1\Rightarrow n>1+\frac{1}{n}.$$

This inequality holds for all $n>2$. Thus, the sequence $x_n=-\frac{n^2}{2^n}$ starting from the third term monotonically increases, and the inequality $-\frac{9}{8}\le x_n<0$ holds for all terms of the sequence.

The smallest term of the sequence is $x_3=-\frac{9}{8}$.

Answer: $x_3=-\frac{9}{8}$.

7. Using logical symbols, write the following statements, as well as their negations:

a) The sequence is bounded;

Solution.

The sequence is bounded:

$$\mathrm{\exists }A>0;\mathrm{\forall }n\in N(|x_n|\le A).$$

The negation is: The sequence is unbounded.

$$\mathrm{\forall }A>0;\mathrm{\exists }n\in N(|x_n|>A).$$

Answer: $\mathrm{\exists }A>0;\mathrm{\forall }n\in N(|x_n|\le A);$ $\mathrm{\forall }A>0;\mathrm{\exists }n\in N(|x_n|>A).$

b) The sequence is monotonically increasing.

Solution:

The sequence is monotonically increasing:

$$\mathrm{\forall }n\in N(x_n<x_{n+1}).$$

The negation:

$$\mathrm{\exists }n\in N(x_n\ge x_{n+1}).$$

Answer: $\mathrm{\forall }n\in N(x_n<x_{n+1});$ $\mathrm{\exists }n\in N(x_n\ge x_{n+1}).$

d) The number $a$ is a limit point of the sequence.

Solution:

The number $a$ is a limit point of the sequence:

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }n\in N(|x_n-a|<\epsilon ).$$

The negation:

$$\mathrm{\exists }\epsilon >0\mathrm{\forall }n\in N(|x_n-a|\ge \epsilon ).$$

Answer: $\mathrm{\forall }\epsilon >0\mathrm{\exists }n\in N(|x_n-a|<\epsilon );$ $\mathrm{\exists }\epsilon >0\mathrm{\forall }n\in N(|x_n-a|\ge \epsilon ).$

1.230. Find $a=\underset{n\to \mathrm{\infty }}{lim}{x}_n$ and determine the number $N(\epsilon )$ such that $|x_n-a|<\epsilon$ for all $n>N(\epsilon ),$ if:

b) $x_n=\frac{\sqrt{n^2+1}}{n},\epsilon =0.005.$

Solution:

$a=\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt{n^2+1}}{n}=\underset{n\to \mathrm{\infty }}{lim}\sqrt{\frac{n^2+1}{n^2}}=\underset{n\to \mathrm{\infty }}{lim}\sqrt{1+\frac{1}{n^2}}=\sqrt{1+0}=1.$

Let's find the number $N(\epsilon )$ such that $|x_n-a|<\epsilon$ for all $n>N(\epsilon ):$

$$|x_n-a|<\epsilon \Rightarrow |\frac{\sqrt{n^2+1}}{n}-1|<0,005\Rightarrow \sqrt{n^2+1}<1,005n\Rightarrow$$

$$\Rightarrow n^2+1<1,010025n^2\Rightarrow 1<0,010025n^2\Rightarrow 1<0,100125n\Rightarrow$$ $$\Rightarrow n>\frac{1}{0,100125}\sim 9,99.$$

Thus, if we choose the number $N(\epsilon )=9$, then for all $n>N=9$ we have $|x_n-1|<0.005$.

Answer: $a=1,,,N=9.$

c) $x_n=\frac{1}{n}\sin \frac{\pi n}{2},\epsilon =0.001.$

Solution:

$a=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sin \frac{\pi n}{2}=0.$

Let's find the number $N(\epsilon )$ such that $|x_n-a|<\epsilon$ for all $n>N(\epsilon ):$

$$|x_n-a|<\epsilon \Rightarrow |\frac{1}{n}\sin \frac{\pi n}{2}|<0,001\Rightarrow |\sin \frac{\pi n}{2}|<0,001n.$$

In the left side of the inequality, $|\sin \frac{\pi n}{2}|$ takes the value $0$ for even $n$ and $1$ for odd $n$, while in the right side, we have elements of a monotonically increasing sequence $0.001n$. In order for the inequality $|\sin \frac{\pi n}{2}|<0.001n$ to hold for all $n>N(\epsilon )$, we need to find the index from which $0.001n>1$.

$$0,001n>1\Rightarrow n>\frac{1}{0,001}=1000.$$

Thus, if we choose the number $N(\epsilon )=999$, then for all $n>N=999$ we have $|x_n|<0.001$.

Answer: $a=0,,,N=999.$

Compute the limits:

8. $\underset{n\to \mathrm{\infty }}{lim}\frac{n-1}{3n}.$

Solution:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{n-1}{3n}=\underset{n\to \mathrm{\infty }}{lim}(\frac{n}{3n}-\frac{1}{3n})=\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{3}-\frac{1}{3n})=\frac{1}{3}.$$

Answer: $\frac{1}{3}.$

9. $\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1{)}^2}{2n^3}.$

Solution.

$$\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1{)}^2}{2n^3}=\underset{n\to \mathrm{\infty }}{lim}\frac{n^2+2n+1}{2n^3}=\underset{n\to \mathrm{\infty }}{lim}(\frac{n^2}{2n^3}+\frac{2n}{2n^3}+\frac{1}{2n^3})=$$ $$=\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{2n}+\frac{2}{2n^2}+\frac{1}{2n^3})=0.$$

Answer: $0.$

10. $\underset{n\to \mathrm{\infty }}{lim}\frac{(n+2{)}^3-(n-2{)}^3}{95n^3+39n}.$

Solution.

$$\underset{n\to \mathrm{\infty }}{lim}\frac{(n+2{)}^3-(n-2{)}^3}{95n^3+39n}=$$ $$=\underset{n\to \mathrm{\infty }}{lim}\frac{n^3+6n^2+12n+8-n^3+6n^2-12n+8}{95n^3+39n}=$$ $$=\underset{n\to \mathrm{\infty }}{lim}\frac{12n^2+16}{95n^3+39n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{12n^2}{n^3}+\frac{16}{n^3}}{\frac{95n^3}{n^3}+\frac{39n}{n^3}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{12}{n}+\frac{16}{n^3}}{95+\frac{39}{n^2}}=\frac{0}{95}=0.$$

Answer: $0.$

11. $\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[3]{n^4+3n+1}}{n-1}.$

Solution.

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[3]{n^4+3n+1}}{n-1}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{\sqrt[3]{n^4+3n+1}}{n^{4/3}}}{\frac{n-1}{n^{4/3}}}=$$ $$=\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[3]{\frac{n^4}{n^4}+\frac{3n}{n^4}+\frac{1}{n^4}}}{\frac{n}{n^{4/3}}-\frac{1}{n^{4/3}}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[3]{1+\frac{3}{n^3}+\frac{1}{n^4}}}{\frac{1}{n^{1/4}}-\frac{1}{n^{4/3}}}=\left[\frac{1}{+0}\right]=+\mathrm{\infty }.$$

Answer: $+\mathrm{\infty }.$

12. $\underset{n\to \mathrm{\infty }}{lim}(\sqrt{n+1}-\sqrt{n}).$

Solution.

$$\underset{n\to \mathrm{\infty }}{lim}(\sqrt{n+1}-\sqrt{n})=\underset{n\to \mathrm{\infty }}{lim}\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=$$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n+1}+\sqrt{n}}=\left[\frac{1}{\mathrm{\infty }}\right]=0.$$

Answer: $0.$

13. $\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}).$

Solution.

$$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2})=\underset{n\to \mathrm{\infty }}{lim}\frac{1+2+...+n-1}{n^2}.$$ In the numerator, we have an arithmetic progression with $a_1=1$ and $d=1$. We can find the sum of the first $n-1$ terms using the formula $S_{n-1}=\frac{a_1+a_{n-1}}{2}(n-1):$

$$S_{n-1}=\frac{1+n-1}{2}(n-1)=\frac{n(n-1)}{2}.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2})=\underset{n\to \mathrm{\infty }}{lim}\frac{n(n-1)}{2n^2}=\underset{n\to \mathrm{\infty }}{lim}\frac{n-1}{2n}=$$

$$=\underset{n\to \mathrm{\infty }}{lim}(\frac{n}{2n}-\frac{1}{2n})=\frac{1}{2}.$$

Answer: $\frac{1}{2}.$

14. Is the given sequence infinitely large $x_n=n^{(-1{)}^n}.$

Solution.

The elements of this sequence with even indices can be written as $x_{2k}=(2k{)}^{(-1{)}^{2k}}=2k;$

with odd indices can be written as $x_{2k+1}=(2k+1{)}^{(-1{)}^{2k+1}}=\frac{1}{2k+1}.$

By the definition, a sequence $(x_n)n\in \mathbb{N}$ is called infinitely large (converging to infinity), denoted as $limn\to \mathrm{\infty }=\mathrm{\infty }$, if for any number $E>0$ there exists a number $N(E)$ such that for $n>N(E)$ the inequality $|x_n|>E.$

For the given sequence, for any fixed number $E>0$, for all $N$, there exists a number $n:$ $n=(2E+1)N>N$ for odd $N$ and $n=(2E+1)(N+1)>N$ for even $N$ such that $x_n=\frac{1}{n}\leq\frac{1}{(2E+1)(2N+1)}

Thus, the given sequence is not infinitely large.

Answer: No.

15. Find all limit points of the sequence $x_n=\cos \frac{\pi n}{4}.$

Solution.

Let's divide this sequence into several subsequences:

$\{x_{8k}\}:x_{8k}=\cos \frac{8k\pi }{4}=\cos 2k\pi =1;$

$\{x_{8k+1}\}:x_{8k+1}=\cos \frac{(8k+1)\pi }{4}=\cos (2k\pi +\frac{\pi }{4})=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}};$

$\{x_{8k+2}\}:x_{8k+2}=\cos \frac{(8k+2)\pi }{4}=\cos (2k\pi +\frac{2\pi }{4})=\cos \frac{2\pi }{4}=\frac{\pi }{2}=0;$

$\{x_{8k+3}\}:x_{8k+3}=\cos \frac{(8k+3)\pi }{4}=\cos (2k\pi +\frac{3\pi }{4})=\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}};$

$\{x_{8k+4}\}:x_{8k+4}=\cos \frac{(8k+4)\pi }{4}=\cos (2k\pi +\frac{4\pi }{4})=\cos \frac{4\pi }{4}=\cos \pi =-1;$

$\{x_{8k+5}\}:x_{8k+5}=\cos \frac{(8k+5)\pi }{4}=\cos (2k\pi +\frac{5\pi }{4})=\cos \frac{5\pi }{4}=-\frac{1}{\sqrt{2}};$

$\{x_{8k+6}\}:x_{8k+6}=\cos \frac{(8k+6)\pi }{4}=\cos (2k\pi +\frac{6\pi }{4})=\cos \frac{6\pi }{4}=\frac{3\pi }{2}=0;$

$\{x_{8k+7}\}:x_{8k+7}=\cos \frac{(8k+7)\pi }{4}=\cos (2k\pi +\frac{7\pi }{4})=\cos \frac{7\pi }{4}=\frac{1}{\sqrt{2}}.$

Thus, all elements of the sequence take only 5 different values: $1,,\frac{1}{\sqrt{2}},,0,,-\frac{1}{\sqrt{2}},,,1,$ and each of these values repeats infinitely many times. Therefore, for any ${\epsilon }_1>0$, there exists an infinitely large number of terms in this sequence (all terms of the subsequence $x_{8k}$), satisfying the condition $|x_n-1|<{\epsilon }_1.$

Thus, $1$ is a limit point of the sequence. Similarly, $\frac{1}{\sqrt{2}},,0,,-\frac{1}{\sqrt{2}},,,1$ are also limit points of the sequence.

Answer: $1,,\frac{1}{\sqrt{2}},,0,,-\frac{1}{\sqrt{2}},,,1.$

In problems 1.255 and 1.257, for the given sequences $(x_n)n\in \mathbb{N}$, find $inf(x_n),,,sup(x_n),,,\bar{\lim }n\to \mathrm{\infty }{x}_n,$ $\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}{x}_n,$

16. $x_n=1+\frac{1}{n}.$

Solution.

Let's write the sequence as

$$X=\{x_n:x_n=1+\frac{1}{n}\}=$$ $$=\{2,1\frac{1}{2},1\frac{1}{3},..,1+\frac{1}{n},...\}=\{2,\frac{3}{2},\frac{4}{3},...,\frac{n+1}{n},...\}.$$

This set has a largest element $M=2$ because for all elements of the sequence $x_n$, the inequality $x_n\le 2$ holds. Also, $x_1=2$.

The given sequence does not have a smallest element, because for any element $x_n=1+\frac{1}{n}$, there always exists an element $x_{n+1}=1+\frac{1}{n+1}\in X$ for which the inequality $x_{n+1}\le x_n$ holds.

Since for all elements of the sequence $x_n$, the inequality $x\le 2$ holds, and $2$ is an element of the sequence, the set of upper bounds for the set $X$ is the set $[2,+\mathrm{\infty })$ with the smallest element being $2$. Thus, $supX=sup(x_n)=2$.

The given sequence does not have a smallest element. It is evident that for all elements $x_n$ of the sequence, $x\ge 1$. Therefore, the set $X$ is bounded from below. We will show that $1$ is a limit point of the set $X$. Indeed, for any $\epsilon >0$, we can find a natural number $$n>\frac{1}{\epsilon }\Rightarrow \frac{1}{n}<\epsilon \Rightarrow 1+\frac{1}{n}-1<\epsilon ,1+\frac{1}{n}\in X.$$ Thus, the set of lower bounds for $X$ is the set $(-\mathrm{\infty },1]$ with the largest element being $1$. Hence, we find $infX=inf(x_n)=1$.

Since the given sequence has a limit,

$$\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}(1+\frac{1}{n})=\underset{n\to \mathrm{\infty }}{\bar{\lim }}(1+\frac{1}{n})=\underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n})=1.$$

Answer: $sup(x_n)=2,$ $inf(x_n)=1;$ $\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}(1+\frac{1}{n})=\underset{n\to \mathrm{\infty }}{\bar{\lim }}(1+\frac{1}{n})=1.$

17. $x_n=(-1{)}^n(2n+1).$

Solution.

Let's express the sequence as

$$X=\{x_n:x_n=(-1{)}^n(2n+1)\}=$$ $$=\{-3,5,-7,9,..,(-1{)}^n(2n+1),...\}.$$

From the given sequence, we can identify two subsequences:

$x_{2k}=4k+1;$

$x_{2k+1}=-(4k+3).$

Let's find the limits of these subsequences as $k\to \mathrm{\infty }:$

$\underset{k\to \mathrm{\infty }}{lim}(4k+1)=+\mathrm{\infty };$

$\underset{k\to \mathrm{\infty }}{lim}(-(4k+3))=-\mathrm{\infty }.$

Thus, the given sequence is unbounded both from above and below.

$\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}{x}_n=$ $\underset{k\to \mathrm{\infty }}{lim}{x}_{2k+1}=$ $\underset{k\to \mathrm{\infty }}{lim}(-(4k+3))=-\mathrm{\infty }.$

$\underset{n\to \mathrm{\infty }}{\bar{\lim }}{x}_n=$ $\underset{k\to \mathrm{\infty }}{lim}{x}_{2k}=$ $\underset{k\to \mathrm{\infty }}{lim}(4k+1)=+\mathrm{\infty }.$

Answer: The sequence is unbounded both from above and below; $\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}{x}_n=-\mathrm{\infty };,,,,\underset{x\to \mathrm{\infty }}{\bar{\lim }}{x}_n=+\mathrm{\infty }.$

Homework:

In problems 1, 2, write the first five terms of the sequence:

1. $x_n=n(1-(-1{)}^n).$

2. $x_n=(-1^n)\arcsin \frac{\sqrt{3}}{2}+\pi n.$

In problems 3 - 5, write the general term of the sequence:

3. $\{2,\frac{4}{3},\frac{6}{5},\frac{8}{7},...\}$

4. $\{-3,\frac{5}{3},-\frac{7}{5},\frac{9}{7},-\frac{11}{9}\}$

5. $\{0,\frac{\sqrt{2}}{2},1,\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2},-1,-\frac{\sqrt{2}}{2},0,...\}.$

In problems 6, 7, find the greatest (smallest) term of the bounded above (below) sequence $(x_n{)}_{n\in \mathbb{N}}$.

6. $x_n=e^{10n-n^2-24}.$

7. $x_n=\frac{\sqrt{n}}{9+n}.$

8. Using logical symbols, write down the following statements as well as their negations:

в) The number $a$ is the limit of the sequence.

9. Find $a=\underset{n\to \mathrm{\infty }}{lim}{x}_n$ and determine the number $N(\epsilon )$ such that $|x_n-a|<\epsilon$ for all $n>N(\epsilon ),$ if:

a) $x_n=0.333333...3$ (the digit 3 appears $n$ times after the decimal point), $\epsilon =0.001.$

b) $x_n=\frac{5n^2+1}{7n^2-3},$ $\epsilon =0.005.$

Compute the limits:

10. $\underset{n\to \mathrm{\infty }}{lim}\frac{5n+1}{7-9n}.$

11. $\underset{n\to \mathrm{\infty }}{lim}\frac{3n^2-7n+1}{2-5n-6n^2}.$

12. $\underset{n\to \mathrm{\infty }}{lim}(\frac{2n-1}{5n+7}-\frac{1+2n^3}{2+5n^3}).$

13. $\underset{n\to \mathrm{\infty }}{lim}{n}^{3/2}(\sqrt{n^3+1}-\sqrt{n^3-2}).$

14. $\underset{n\to \mathrm{\infty }}{lim}\frac{2^n+3^n}{2^n-3^n}.$

15. $\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n(n+1)}).$

In problems 16, 17, determine which of the given sequences are infinitely large.

16. $x_n=2^{\sqrt{n}}.$

17. $x_n=n\sin \frac{\pi n}{2}.$

Find all limit points of the sequence:

18. $x_n=\frac{2+(-1{)}^n}{2-(-1{)}^n}.$

In problems 19 and 20, for the given sequences $(x_n)n\in \mathbb{N}$, find $inf(x_n),,,sup(x_n),,,\bar{\lim }n\to \mathrm{\infty }{x}_n,$ $\underset{n\to \mathrm{\infty }}{\underset{―}{\lim }}{x}_n,$

19. $x_n=\frac{n+1}{n}{\cos }^2\frac{\pi n}{4}.$

20. $x_n=\frac{n+2}{n-2}\sin \frac{\pi n}{3},n\ge 2.$

Tags: Boundedness of sequence, Cauchy Criterion, Limit of a sequence, Sequence, calculus, mathematical analysis