Sequence. Boundedness and monotonicity of a sequence. Limit of a sequence.
A sequence of real numbers is defined as a function $f: \mathbb{N} \rightarrow \mathbb{R}$ 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 $\varepsilon>0$ there exists a number $N(\varepsilon)$ such that for $n>N(\varepsilon)$ the inequality $|x{n+p}-x_n|<\varepsilon$ 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 $\varepsilon>0$ there exists a number $N(\varepsilon)$ such that for $n>N(\varepsilon)$ the inequality $|x{n+p}-x_n|<\varepsilon$ holds for any $p\in \mathbb{N}$.
A sequence $(x_n){n\in \mathbb{N}}$ is called infinitesimal if $\lim\limits{n\rightarrow\infty}=0$.
A sequence $(x_n){n\in \mathbb{N}}$ is called infinitely large (converging to infinity), denoted as $\lim\limits{n\rightarrow\infty}=\infty$, 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
$$\lim\limits_{n\rightarrow\infty}=+\infty\qquad \lim\limits_{n\rightarrow\infty}=-\infty$$
The number $a$ is called a limit point of the sequence $(x_n)_{n\in \mathbb{N}}$ if for any $\varepsilon>0$ there exist infinitely many terms of this sequence satisfying the condition $|x_n-a|<\varepsilon$.
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
$\varlimsup\limits_{n\rightarrow\infty}x_n$ (${\varliminf\limits_{n\rightarrow\infty}}x_n$).
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 $n\geq 6,$ 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 $n \geq 3$, 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 $-\frac{9}{8}\leq x_n<0$ 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:
$$\exists A>0;\,\,\, \forall n\in N (|x_n|\leq A).$$
The negation is: The sequence is unbounded.
$$\forall A>0;\,\,\, \exists n\in N (|x_n|>A).$$
Answer: $\exists A>0;\,\,\, \forall n\in N (|x_n|\leq A);$ $\forall A>0;\,\,\, \exists n\in N (|x_n|>A).$
b) The sequence is monotonically increasing.
Solution:
The sequence is monotonically increasing:
$$ \forall n\in N (x_n< x_{n+1}).$$
The negation:
$$\exists n\in N (x_n\geq x_{n+1}).$$
Answer: $ \forall n\in N (x_n< x_{n+1});$ $\exists n\in N (x_n\geq x_{n+1}).$
d) The number $a$ is a limit point of the sequence.
Solution:
The number $a$ is a limit point of the sequence:
$$\forall\varepsilon>0 \exists n\in N (|x_n-a|< \varepsilon).$$
The negation:
$$\exists\varepsilon>0 \forall n\in N (|x_n-a|\geq \varepsilon).$$
Answer: $\forall\varepsilon>0 \exists n\in N (|x_n-a|< \varepsilon);$ $\exists\varepsilon>0 \forall n\in N (|x_n-a|\geq \varepsilon).$
1.230. Find $a=\lim\limits_{n\rightarrow \infty}x_n$ and determine the number $N(\varepsilon)$ such that $|x_n-a|<\varepsilon$ for all $n>N(\varepsilon),$ if:
b) $x_n=\frac{\sqrt{n^2+1}}{n},\qquad \varepsilon=0.005.$
Solution:
$a=\lim\limits_{n\rightarrow\infty}\frac{\sqrt{n^2+1}}{n}=\lim\limits_{n\rightarrow \infty}\sqrt{\frac{n^2+1}{n^2}}=\lim\limits_{n\rightarrow\infty}\sqrt{1+\frac{1}{n^2}}=\sqrt{1+0}=1.$
Let's find the number $N(\varepsilon)$ such that $|x_n-a|<\varepsilon$ for all $n>N(\varepsilon):$
$$|x_n-a|<\varepsilon\Rightarrow \left|\frac{\sqrt{n^2+1}}{n}-1\right|<0,005\Rightarrow \sqrt{n^2+1}<1,005n\Rightarrow $$
$$\Rightarrow n^2+1<1,010025n^2\Rightarrow 1<0,010025n^2\Rightarrow 1<0,100125n\Rightarrow $$ $$\Rightarrow n>\frac{1}{0,100125}\sim 9,99.$$
Thus, if we choose the number $N(\varepsilon)=9$, then for all $n>N=9$ we have $|x_n-1|<0.005$.
Answer: $a=1,,, N=9.$
c) $x_n=\frac{1}{n}\sin\frac{\pi n}{2},\qquad \varepsilon=0.001.$
Solution:
$a=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sin\frac{\pi n}{2}=0.$
Let's find the number $N(\varepsilon)$ such that $|x_n-a|<\varepsilon$ for all $n>N(\varepsilon):$
$$|x_n-a|<\varepsilon\Rightarrow \left|\frac{1}{n}\sin\frac{\pi n}{2}\right|<0,001\Rightarrow |\sin\frac{\pi n}{2}|<0,001 n.$$
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 $n>N(\varepsilon)$, 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 $N(\varepsilon)=999$, then for all $n>N=999$ we have $|x_n|<0.001$.
Answer: $a=0,,, N=999.$
Compute the limits:
8. $\lim\limits_{n\rightarrow \infty}\frac{n-1}{3n}.$
Solution:
$$\lim\limits_{n\rightarrow \infty}\frac{n-1}{3n}=\lim\limits_{n\rightarrow \infty}\left(\frac{n}{3n}-\frac{1}{3n}\right)=\lim\limits_{n\rightarrow \infty}\left(\frac{1}{3}-\frac{1}{3n}\right)=\frac{1}{3}.$$
Answer: $\frac{1}{3}.$
9. $\lim\limits_{n\rightarrow \infty}\frac{(n+1)^2}{2n^3}.$
Solution.
$$\lim\limits_{n\rightarrow \infty}\frac{(n+1)^2}{2n^3}=\lim\limits_{n\rightarrow \infty}\frac{n^2+2n+1}{2n^3}=\lim\limits_{n\rightarrow \infty}\left(\frac{n^2}{2n^3}+\frac{2n}{2n^3}+\frac{1}{2n^3}\right)=$$ $$=\lim\limits_{n\rightarrow \infty}\left(\frac{1}{2n}+\frac{2}{2n^2}+\frac{1}{2n^3}\right)=0.$$
Answer: $0.$
10. $\lim\limits_{n\rightarrow \infty}\frac{(n+2)^3-(n-2)^3}{95n^3+39n}.$
Solution.
$$\lim\limits_{n\rightarrow \infty}\frac{(n+2)^3-(n-2)^3}{95n^3+39n}=$$ $$=\lim\limits_{n\rightarrow \infty}\frac{n^3+6n^2+12n+8-n^3+6n^2-12n+8}{95n^3+39n}=$$ $$=\lim\limits_{n\rightarrow \infty}\frac{12n^2+16}{95n^3+39n}=\lim\limits_{n\rightarrow \infty}\frac{\frac{12n^2}{n^3}+\frac{16}{n^3}}{\frac{95n^3}{n^3}+\frac{39n}{n^3}}=\lim\limits_{n\rightarrow \infty}\frac{\frac{12}{n}+\frac{16}{n^3}}{95+\frac{39}{n^2}}=\frac{0}{95}=0.$$
Answer: $0.$
11. $\lim\limits_{n\rightarrow \infty}\frac{\sqrt[3]{n^4+3n+1}}{n-1}.$
Solution.
$$\lim\limits_{n\rightarrow\infty}\frac{\sqrt[3]{n^4+3n+1}}{n-1}=\lim\limits_{n\rightarrow\infty}\frac{\frac{\sqrt[3]{n^4+3n+1}}{n^{4/3}}}{\frac{n-1}{n^{4/3}}}=$$ $$=\lim\limits_{n\rightarrow \infty}\frac{\sqrt[3]{\frac{n^4}{n^4}+\frac{3n}{n^4}+\frac{1}{n^4}}}{\frac{n}{n^{4/3}}-\frac{1}{n^{4/3}}}=\lim\limits_{n\rightarrow \infty}\frac{\sqrt[3]{1+\frac{3}{n^3}+\frac{1}{n^4}}}{\frac{1}{n^{1/4}}-\frac{1}{n^{4/3}}}=\left[\frac{1}{+0}\right]=+\infty.$$
Answer: $+\infty.$
12. $\lim\limits_{n\rightarrow \infty}(\sqrt{n+1}-\sqrt n).$
Solution.
$$\lim\limits_{n\rightarrow \infty}(\sqrt{n+1}-\sqrt n)=\lim\limits_{n\rightarrow\infty}\frac{(\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n)}{\sqrt{n+1}+\sqrt n}=$$
$$\lim\limits_{n\rightarrow\infty}\frac{n+1-n}{\sqrt{n+1}+\sqrt n}=\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}+\sqrt n}=\left[\frac{1}{\infty}\right]=0.$$
Answer: $0.$
13. $\lim\limits_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right).$
Solution.
$$\lim\limits_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right)=\lim\limits_{n\rightarrow\infty}\frac{1+2+...+n-1}{n^2}.$$ 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,
$$\lim\limits_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right)=\lim\limits_{n\rightarrow\infty}\frac{n(n-1)}{2n^2}=\lim\limits_{n\rightarrow\infty}\frac{n-1}{2n}=$$
$$=\lim\limits_{n\rightarrow\infty}\left(\frac{n}{2n}-\frac{1}{2n}\right)=\frac{1}{2}.$$
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 $\lim\limits{n\rightarrow\infty}=\infty$, 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 $\varepsilon_1>0$, there exists an infinitely large number of terms in this sequence (all terms of the subsequence ${x_{8k}}$), satisfying the condition $|x_n-1|<\varepsilon_1.$
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),,, \varlimsup\limits{n\rightarrow\infty}x_n, $ $\varliminf\limits_{n\rightarrow\infty}x_n, $
16. $x_n=1+\frac{1}{n}.$
Solution.
Let's write the sequence as
$$X=\left\{x_n: x_n=1+\frac{1}{n}\right\}=$$ $$=\left\{2,\, 1\frac{1}{2},\, 1\frac{1}{3}, .., 1+\frac{1}{n},...\right\}=\left\{2,\, \frac{3}{2},\,\,\frac{4}{3},\,\,...,\frac{n+1}{n},...\right\}.$$
This set has a largest element $M=2$ because for all elements of the sequence ${x_n}$, the inequality $x_n\leq 2$ 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 $x_{n+1}\leq x_n$ holds.
Since for all elements of the sequence $x_n$, the inequality $x\leq 2$ holds, and $2$ is an element of the sequence, the set of upper bounds for the set $X$ is the set $\left[2, +\infty\right)$ with the smallest element being $2$. Thus, $\sup X=\sup (x_n)=2$.
The given sequence does not have a smallest element. It is evident that for all elements $x_n$ of the sequence, $x\geq 1$. Therefore, the set $X$ is bounded from below. We will show that $1$ is a limit point of the set $X$. Indeed, for any $\varepsilon>0$, we can find a natural number $$n>\frac{1}{\varepsilon}\,\,\Rightarrow \frac{1}{n}<\varepsilon\Rightarrow\,\,1+\frac{1}{n}-1<\varepsilon,\quad 1+\frac{1}{n}\in X.$$ Thus, the set of lower bounds for $X$ is the set $(-\infty, 1]$ with the largest element being $1$. Hence, we find $\inf X=\inf(x_n)=1$.
Since the given sequence has a limit,
$$\varliminf\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)=\varlimsup\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)=\lim\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)=1.$$
Answer: $\sup (x_n)=2,$ $\inf (x_n)=1;$ $\varliminf\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)=\varlimsup\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)=1.$
17. $x_n=(-1)^n(2n+1).$
Solution.
Let's express the sequence as
$$X=\left\{x_n: x_n=(-1)^n(2n+1)\right\}=$$ $$=\left\{-3,\, 5,\, -7,\, 9, .., (-1)^n(2n+1),...\right\}.$$
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 $k\rightarrow\infty:$
$\lim\limits_{k\rightarrow\infty}(4k+1)=+\infty;$
$\lim\limits_{k\rightarrow\infty}(-(4k+3))=-\infty.$
Thus, the given sequence is unbounded both from above and below.
$\varliminf\limits_{n\rightarrow\infty}x_n=$ $\lim\limits_{k\rightarrow\infty}x_{2k+1}=$ $\lim\limits_{k\rightarrow\infty}(-(4k+3))=-\infty.$
$\varlimsup\limits_{n\rightarrow\infty}x_n=$ $\lim\limits_{k\rightarrow\infty}x_{2k}=$ $\lim\limits_{k\rightarrow\infty}(4k+1)=+\infty.$
Answer: The sequence is unbounded both from above and below; $\varliminf\limits_{n\rightarrow\infty}x_n=-\infty;,,,,\varlimsup\limits_{x\rightarrow\infty}x_n=+\infty.$
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. $\left\{2,\, \frac{4}{3},\,\,\frac{6}{5},\,\, \frac{8}{7}, ...\right\}$
4. $\left\{-3,\,\,\frac{5}{3},\,\,-\frac{7}{5},\,\,\frac{9}{7},\,\,-\frac{11}{9}\right\}$
5. $\left\{0,\,\,\frac{\sqrt 2}{2},\,\,1,\,\,\frac{\sqrt 2}{2},\,\,0,\,\,-\frac{\sqrt 2}{2}, -1,\, -\frac{\sqrt 2}{2}, 0, ...\right\}.$
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 $a=\lim\limits_{n\rightarrow \infty}x_n$ and determine the number $N(\varepsilon)$ such that $|x_n-a|<\varepsilon$ for all $n>N(\varepsilon),$ if:
a) $x_n=0.333333...3$ (the digit 3 appears $n$ times after the decimal point), $\varepsilon=0.001.$
b) $x_n=\frac{5n^2+1}{7n^2-3},$ $\varepsilon=0.005.$
Compute the limits:
10. $\lim\limits_{n\rightarrow \infty}\frac{5n+1}{7-9n}.$
11. $\lim\limits_{n\rightarrow \infty}\frac{3n^2-7n+1}{2-5n-6n^2}.$
12. $\lim\limits_{n\rightarrow \infty}\left(\frac{2n-1}{5n+7}-\frac{1+2n^3}{2+5n^3}\right).$
13. $\lim\limits_{n\rightarrow \infty}n^{3/2}(\sqrt{n^3+1}-\sqrt {n^3-2}).$
14. $\lim\limits_{n\rightarrow \infty}\frac{2^n+3^n}{2^n-3^n}.$
15. $\lim\limits_{n\rightarrow \infty}\left(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n(n+1)}\right).$
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),,, \varlimsup\limits{n\rightarrow\infty}x_n, $ $\varliminf\limits_{n\rightarrow\infty}x_n, $
19. $x_n=\frac{n+1}{n}\cos^2\frac{\pi n}{4}.$
20. $x_n=\frac{n+2}{n-2}\sin\frac{\pi n}{3},\quad n\geq 2.$
