Limit of Sequence

Sequences For each number of $1,2,3,\ldots,n,\ldots$ , there corresponds unique real number

$\displaystyle a_{1},a_{2},a_{3},\ldots,a_{n},\ldots$

We call this list of numbers sequence and call $a_{1},a_{2},\ldots$ term.

NOTE Once the $n$ th term is found, all terms can be derived. Thus we call the $n$ th term general term.

Arithmetic sequence A sequence $\{a_{n}\}$ whose difference between the consecutive terms is constant is called Arithmetic sequence. Note that we can express $a_{n} = (a_{n} - a_{n-1}) + (a_{n-1} - a_{n-2}) + \cdots + (a_{2}- a_{1}) + a_{1} = (n-1)d + a_{1}$ . Geometric sequence A sequence $\{a_{n}\}$ whose ratio of consecutive terms is constant is called geometric sequence. Note that $a_{n} = \frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} + \cdots + \frac{a_{2}}{a_{1}} \cdot a_{1} = a_{1}r^{n-1}$ .

Bounded sequence A sequence $\{a_{n}\}$ is called bounded above if there exists a number $M$ such that $a_{n} \leq M$ for all $n$ . A sequence $\{a_{n}\}$ is called bounded below if there exists a number $m$ such that $a_{n} \geq m$ for all $n$ . Furthermore, a sequence $\{a_{n}\}$ is called bounded if it is bounded above and bounded below..

NOTE The sequence $\{a_{n}\} = \{\frac{1}{n}\}$ is bounded. For $\frac{1}{n} > 0$ and $\frac{1}{n} < 1$ for all $n$ .

Monotonicity If $a_{n+1} \geq a_{n}$ for all $n$ , a sequence $\{a_{n}\}$ is called monotonically increasing. If $a_{n+1} \leq a_{n}$ for all $n$ , a sequence $\{a_{n}\}$ is called monotonicall decreasing sequence.

Strictly increasing As $n$ gets larger, the value of $a_{n}$ gets larger without equality. Then we say strictly increasing.

Example 1..14 Determine the sequence $\displaystyle{\{a_{n}\} = \{\frac{n}{n+1}\}}$ is monotonically increasings.

SOLUTION We use the ratio to check to see. Note that $a_{n+1} = \frac{n+1}{n+2}$ .

$a_{n+1} = \frac{n+1}{(n+1)+1} = \frac{n+1}{n+2}$ .
$n^2 + n + 1 > n^2 + n$ implies $\frac{n^2+n+1}{n^2 + n} > 1$ .

$\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{n+1}{n+2}\cdot \frac{n+1}{n} = \frac{n^{2}+2n+1}{n^{2}+2} > 1$

Thus $a_{n+1} \geq a_{n}$ and $\{a_{n}\}$ is monotonically increasing $\blacksquare$

Exercise 1..14 Determine whether the following sequence is increasing or decreasing.

$\displaystyle a_{1} = 1, a_{n+1} = \sqrt{3a_{n}}$

SOLUTION Since $a_{1} = 1, a_{2} = \sqrt{3a_{1}} = \sqrt{3}$ , we have $a_{1} < a_{2}$ . Now use mathematical induction. Show the statement is true for $n=1$

and assume true $n=k$

. If the statement is true for $n=k+1$

, then it is ture for all $n$

For $n=1$ , $a_{n} < a_{n+1}$ is true. Now assume that $a_{k} < a_{k+1}$ is true. Then we have

$\displaystyle a_{k+1} = \sqrt{3a_{k}} < \sqrt{3a_{k+1}} = a_{k+2}$

Thus by mathematical induction, $a_{n} < a_{n+1}$ for all $n$

$\blacksquare$

Limit of sequence As $n$ approaches $\infty$ , $a_{n}$ approaches $a$ . Then we write

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = a$

We say the sequence $\{a_{n}\}$ converges to $a$

and call this $a$

limit.

NOTE When a sequence converges to $a$ , $a$ has to be a real number. Thus we can not use $+\infty$ or $-\infty$ for $a$ . When a sequence $\{a_{n}\}$ does not converge, we say $\{a_{n}\}$ diverges.

Understanding Limit We can think of $\lim_{n \to \infty}a_{n} = a$ as the value of $\vert a_{n} - a\vert$ gets close to 0.

Types of divergence There are basically two types of divergence. As $n$ gets larger, $a_{n}$ gets larger without bound. In this case, we write $\lim_{n \to \infty}a_{n} = \infty$ . As $n$ gets larger, $a_{n}$ takes the postive value and negative value and never approaches any number. In this case, we say the sequence bibrates.

Limit properties

Theorem 1..1 Suppose that $\displaystyle{\lim_{n \rightarrow \infty}a_{n} = a, \lim_{n \rightarrow \infty}b_{n} = b}$ and $c$

is constant. Then we have the followings:

$\displaystyle{1. \lim_{n \rightarrow \infty}(a_{n} \pm b_{n})= a \pm b}$

$\displaystyle{2. \lim_{n \rightarrow \infty}(ca_{n}) = ca }$

$\displaystyle{3. \lim_{n \rightarrow \infty}a_{n}b_{n} = ab}$

$\displaystyle{4. \lim_{n \rightarrow \infty}\frac{a_{n}}{b_{n}} = \frac{a}{b} \mbox{\rm provided} b \neq 0}$

NOTE When the limit exits, four arithmetic operations hold. Express 1. by

$\displaystyle \lim_{n \rightarrow \infty}(a_{n} \pm b_{n}) = \lim_{n \rightarrow \infty}a_{n} \pm \lim_{n \rightarrow \infty} b_{n}$

Then we can memorize by saying "the limit of a sum is the sum of limits".

Limit properties

Theorem 1..2 Suppose that $\displaystyle{\lim_{n \to \infty}a_{n} = \infty, \lim_{n \to \infty}b_{n} = b > 0}$ . Then we have the followings:

$\displaystyle{1. \lim_{n \to \infty}\frac{a_{n}}{b_{n}} = \infty}$

$\displaystyle{2. \lim_{n \to \infty}\frac{b_{n}}{a_{n}} = 0}$

NOTE Suppose that the denominator of the sequence approaches some positive constant as the numerator approaches $\infty$ , Then the sequence gets larger without bound. Thus the sequence diverges. Suppose next that the denominator of the sequence approaches $\infty$ as the numerator approaches a positive constant. Then the limit of the sequence is 0.

Indeterminate What happens if the numerator and the denominator approach $\infty$ . In this case, we say indeterminate. If this happens, we factor by taking out the highest power of $n$ from the numerator and the denominato. Then apply theorem1.1.4.

Example 1..15 Find the limit of the following.

$\displaystyle \{a_{n}\} = \{\frac{3n^2 - 5n}{5n^2 + 2n - 6}\}$

Basic limit $\displaystyle{\lim_{n \to \infty}\frac{1}{n} = 0}$

SOLUTION As $n \rightarrow \infty$ , $\displaystyle{3n^2 - 5n = n^2 (3 - \frac{5}{n})}$ $\longrightarrow \infty$ and $5n^2 + 2n - 6 =$ $\displaystyle{n^2 (5 + \frac{2}{n} - \frac{6}{n^2}) \longrightarrow \infty}$ . Then we factor by taking out $n^2$ to have

$\displaystyle \lim_{n \rightarrow \infty}\frac{3n^2 - 5n}{5n^2 + 2n - 6}$	$\displaystyle =$	$\displaystyle \lim_{n \rightarrow \infty}\frac{n^2(3 - 5/n)}{n^2(5 + 3/n - 6/n^2)}$
	$\displaystyle =$	$\displaystyle \lim_{n \rightarrow \infty}\frac{3 - 5/n}{5 + 3/n - 6/n^2} = \frac{3}{5}\ensuremath{ \blacksquare}$

Exercise 1..15 Find the limit of the sequence $\displaystyle{\{\sqrt{n+1} - \sqrt{n}\}}$

SOLUTION

Rationalization $\begin{displaymath}\begin{array}{ll} & \sqrt{a} - \sqrt{b}\\ &= \frac{(\sqrt{a}... ...+ \sqrt{b}}\\ &= \frac{a - b}{\sqrt{a} + \sqrt{b}} \end{array}\end{displaymath}$

$\displaystyle \lim_{n \to \infty}(\sqrt{n+1} - \sqrt{n})$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{n+1 -n}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty}\frac{1}{\sqrt{n+1} + \sqrt{n}}= 0 \ensuremath{ \blacksquare}$

To find the limit, the theorem above is not enough. For example, consider $\displaystyle{\lim_{n \rightarrow \infty}\frac{\sin{n\theta}}{n}}$ as $\theta \neq 0$ .

**Figure:** Graph of $\sin{n\theta}$
$\includegraphics[width=3cm]{SOFTFIG-1/sinnx_gr1.eps}$

In this problem, we can not obtain the limit by using the theorem 1.1 and the theorem 1.2. To find the limit of $\sin{n\theta}$ , it is useful to use the following theorem. Squeezing theorem

Theorem 1..3 If there exists a number $n_0$

so that $a_{n} < c_{n} < b_{n}$ is true for all $n > n_0$

and

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = \lim_{n \rightarrow \infty}b_{n} = a$

Then $\lim_{n \rightarrow \infty}c_{n} = a$ .

Absolute-value inequalities Subtract $a$ from the inequality $a_n < c_n < b_n$ . Then $a_n -a < c_n -a < b_n -a$ . If $\vert a_n -a\vert < \vert c_n -a\vert$ , then we have $a_n - a > 0$ and $c_n -a > 0$ . Thus $\vert c_n - a\vert < \vert b_n -a\vert$ .

Proof. Since $a_{n} < c_{n} < b_{n}$ , we have either $0 \leq \vert c_{n} - a\vert < \vert a_{n} - a\vert$ or $0 \leq \vert c_{n} - a\vert <\vert b_{n} - a\vert$ . Note that

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = \lim_{n \rightarrow \infty}b_{n} = a$

Thus $\lim_{n \rightarrow \infty}c_{n} = a$ $\blacksquare$

Example 1..16 For $\theta \neq 0$ , find the limit of the following.

$\displaystyle \{a_{n}\} = \{\frac{1}{n}\sin{n\theta}\}$

SOLUTION Since $\vert\sin{n \theta}\vert \leq 1$ , we have

$\displaystyle 0 \leq \vert\frac{\sin{n \theta}}{n}\vert \leq \frac{1}{n}$

for all $n, \theta$ . Thus we can sandwich $\displaystyle{\vert\frac{\sin{n \theta}}{n}\vert}$ using 0 and $\displaystyle{\frac{1}{n}}$ . Now taking the limit of 0 and $\displaystyle{\frac{1}{n}}$ , we obtain

$\displaystyle \lim_{n \rightarrow \infty} 0 = 0, \lim_{n \rightarrow \infty} \frac{1}{n} = 0.$

Thus by the squeezing theorem,

$\displaystyle \lim_{n \rightarrow \infty}\vert\frac{\sin{n \theta}}{n}\vert = 0.$

Note that

$\displaystyle \vert\frac{\sin{n \theta}}{n} - 0\vert = \vert\vert\frac{\sin{n \theta}}{n}\vert - 0 \vert$

we have

$\displaystyle \lim_{n \rightarrow \infty}\frac{\sin{n \theta}}{n} = 0\ensuremath{ \blacksquare}$

Property of $\sin{n\theta}$ Since $\sin{n\theta}$ is a periodic and $0 \leq \sin{n\theta} \leq 1$ .
$\vert\frac{\sin{n\theta}}{n} - 0\vert = \vert\frac{\sin{n\theta}}{n}\vert = \vert\vert\frac{\sin{n\theta}}{n}\vert - 0\vert$

Exercise 1..16 Find the following limit.
1. $\displaystyle{\lim_{n \to \infty}\frac{(-2)^n}{1 + 3^n}}$ 2. $\displaystyle{\lim_{n \to \infty}\frac{5^{n+1} + 4^n}{3^{n-1} - 5^n}}$

SOLUTION 1. $\displaystyle{\lim_{n \to \infty}\frac{(-2)^n}{1 + 3^n} = \lim_{n \to \infty}\f... ...rac{-2}{3}\big)^n}{\big(\frac{1}{3}\big)^n + 1} = 0\ensuremath{ \blacksquare}}$

limit of $\vert r\vert < 1$ implies $\lim_{n \to \infty}r^n = 0$

2. $\displaystyle{\lim_{n \to \infty}\frac{5^{n+1} + 4^n}{3^{n-1} - 5^n} = \lim_{n ... ...big(\frac{3}{5}\big)^{n-1} - 1} = \frac{5}{-1} = -5\ensuremath{ \blacksquare}}$

The limit of $\{r^n\}$ is the base of convergence and divergence.

Bernoulli inequality For $x > 0$ and $n > 1$ ,

$\displaystyle x^{n} \geq n(x - 1) + 1$

Proof $x^{n}-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$ .
For $x > 1$ , we have $(x-1) > 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 > 1 + 1 + \cdots + 1 = n$
For $x = 1$ , we have $(x-1) = 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 = 1 + 1 + \cdots + 1 = n$
For $0 < x < 1$ , we have $(x-1) < 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 < 1 + 1+ \cdots + 1 = n$ . Thus,

$\displaystyle x^{n}-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+x+1) \geq n(x-1) \ensuremath{ \blacksquare}$

Example 1..17 For $\displaystyle{a_{n} = \frac{r^{n+1}}{1 + r^n} (r \neq -1)}$ , find $\lim_{n \to \infty}a_{n}$ .

SOLUTION We first find the limit of $\{r^n\}$ .

Bernoulli inequality By the Bernoulli inequality, $r > 1$ implies that $r^n \geq n(r-1) + 1 \to \infty$ .
For $\vert r\vert < 1$ , we consider $\frac{1}{\vert r\vert}$ . Then $\frac{1}{\vert r\vert} > 1$ and $(\frac{1}{\vert r\vert})^n \geq n(\frac{1}{\vert r\vert} -1) + 1 \to \infty$ .
Thus, for $\vert r\vert < 1$ we have $r^n \to 0$ .

Divide the numerator and the denominator of $a_{n} = \frac{r^{n+1}}{1 + r^n}$ by $r^n$ .

$\displaystyle \lim_{n \rightarrow \infty}r^{n} = \left\{\begin{array}{ll} \inft... ...\ 0 & (\vert r\vert < 1)\\ \mbox{divergence} & (r \leq -1) \end{array}\right.$

Now for

, $a_{n} = \frac{1}{2}$ and $\displaystyle{\lim_{n \to \infty}a _{n} = \frac{1}{2}}$ . For $\vert r\vert < 1$ , $\lim_{n \to \infty}r^n = \lim_{n \to \infty}r^{n+1} = 0$ and $\lim_{n \to \infty}a_{n} = 0$ .

For $r > 1, r < -1$ ,

$\displaystyle \lim_{n \to \infty}a_{n} = \lim_{n \to \infty}\frac{r}{\left(\frac{1}{r}\right)^n + 1} = r\ensuremath{ \blacksquare}$

Exercise 1..17 Find the sequence $\displaystyle{\{a_{n}\} = \{(1 + \frac{1}{n})^{n}\}}$ is monotonically increasing.

Comparison between $a_{n}$ and $a_{n-1}$ .

SOLUTION Since $\displaystyle{\{a_{n}\} = \{(1 + \frac{1}{n})^{n}\}}$ , we have

$\displaystyle a_{n} = (1 + \frac{1}{n})^{n}= (\frac{n+1}{n})^{n}, a_{n-1} = (1+\frac{1}{n-1})^{n-1} = (\frac{n}{n-1})^{n-1}.$

Then,

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

Thus, $a_{n} \geq a_{n-1}$ and $\{a_{n}\}$ is monotonically increasing $\blacksquare$

$(\frac{n-1}{n})^{-1} = \frac{n}{n-1}.$ Bernoulli inequality $(\frac{n^2-1}{n^2})^n \geq n(\frac{n^2-1}{n^2} - 1) + 1 = \frac{n}{n-1}(\frac{n+n^2-1-n}{n})$

Monotone convergence theorem

Theorem 1..4 Every increasing sequence that is bounded above converges. Every decreasing sequence that is bounded below converges..

How to use MCT If you have no idea about the sequence converges or not, this theorem may help.

NOTE If a sequence $\{a_{n}\}$ is monotonically increasing and bounded above, then there exists a number for which the sequence $\{a_{n}\}$ can not become greater than that number. Among all those numbers, we let the least number be $s$ . Then the difference between $s$ and $a_{n}$ becomes small. Thus the sequence $\{a_{n}\}$ converges.

Example 1..18 Given that $\displaystyle{a_{n+1} = \sqrt{3a_{n}}, a_{1} = 1}$ . Determine the sequence $\{a_{n}\}$ converges or diverges..

SOLUTION We show by induction that the sequence $\{a_{n}\}$ converges.

To show the sequence converges, it is enough to show the sequence $\{a_{n}\}$ is bounded above increasing sequence or bounded below decreasing sequence.

By Exercise1.14, we know $\{a_{n}\}$ is monotonically increasing sequence. So, we need to show the sequence is bounded above. We use mathematical induction on $n$ .

Since $a_{1} = 1 < \sqrt{3} < 3$ , it is true for $n=1$ .

Assume that $a_{k} < 3$ . Then $a_{k+1} = \sqrt{3a_{k}} < \sqrt{3\cdot3} = 3$ . Thus for all $n$ , $a_{n} < 3$ and $\{a_{n}\}$ is bounded above increasing sequence. Therefore, $\{a_{n}\}$ converges $\blacksquare$

Exercise 1..18 Given $\displaystyle{a_{n} = (1 + \frac{1}{n})^n}$ . Determine the sequence $\{a_{n}\}$ converges or diverges.

SOLUTION We have shown in Exercise1.17 that $\displaystyle{\{(1 + \frac{1}{n})^n\}}$ is monotonically increasing. Thus we need to show the sequence is bounded above.

Expand $\displaystyle{(1 + \frac{1}{n})^n}$

$\displaystyle (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n-1}ab^{n-1} + b^{n}.$

is called binomial theorem and $\displaystyle{\binom{n}{j}}$ is called binomial coefficient. using binomial theorem,

$\displaystyle a_{n}$	$\displaystyle =$	$\displaystyle \left(1 + \frac{1}{n}\right)^{n} = \binom{n}{0} \left(\frac{1}{n}... ...binom{n}{1} \left(\frac{1}{n} \right) + \binom{n}{2} \left(\frac{1}{n}\right)^2$
	$\displaystyle +$	$\displaystyle \cdots + \binom{n}{n} \left(\frac{1}{n}\right)^n$
	$\displaystyle =$	$\displaystyle 1 + n\cdot \frac{1}{n} + \frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)^n$
	$\displaystyle <$	$\displaystyle 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots < 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \leq 3.$

Check
$\frac{n(n-1)}{2!}(\frac{1}{n})^2 = \frac{n(n-1)}{2!n\cdot n} < \frac{1}{2!}$ $\frac{1}{3!} = \frac{1}{3\cdot 2} < \frac{1}{2^2}$ $\frac{1}{4!} = \frac{1}{4\cdot 3\cdot 2} < \frac{1}{2^3}$ Thus, the sequence $\{a_{n}\}$ is bounded above increasing sequnce and converges

Euler e

$\displaystyle e = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n$

Eulre The number $e$ is defined by Swiss mathematitian Leonard Euler (1707-1783). The choice of $e$ is from his name,
$\begin{displaymath}\begin{array}{l} e = 2.7182818284590\ldots \end{array}\end{displaymath}$ is known to be irrational.

Example 1..19 Find the limit of the following.

$\displaystyle \lim_{n \to \infty}(1 + \frac{2}{n})^{n}$

SOLUTION Put $\frac{2}{n} = \frac{1}{m}$ . Then $m = \frac{n}{2}$ and $n \to \infty$ implies that $m \to \infty$ . Thus,

$\displaystyle \lim_{n \to \infty}(1 + \frac{2}{n})^{n} = \lim_{m \to \infty}(1 ... ...o \infty}\big((1 + \frac{1}{m})^{m}\big)^{2} = e^{2}\ensuremath{ \blacksquare}$

To use the formula, we wright $\frac{1}{m}$ instead of $\frac{2}{n}$ .

Exercise 1..19 Find the limit of the following.

$\displaystyle \lim_{n \to \infty}(1 - \frac{1}{n})^{n}$

Express $(1-\frac{1}{n}) = \frac{n-1}{n}$ and use $\lim_{m \to \infty}(1 + \frac{1}{m})^{m} = e$ .

SOLUTION Since $1 - \frac{1}{n} = \frac{n-1}{n} = \frac{1}{\frac{n}{n-1}} = \frac{1}{1 + \frac{1}{n-1}}$ , we have

$\displaystyle \lim_{n \to \infty}(1 - \frac{1}{n})^{n}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}(\frac{1}{1 + \frac{1}{n-1}})^{n} = \lim_{n \to \infty}\frac{1}{(1 + \frac{1}{n-1})^{n-1}(1+ \frac{1}{n-1})}$
	$\displaystyle =$	$\displaystyle \frac{1}{e}\ensuremath{ \blacksquare}$

$\binom{n}{j} = \frac{n!}{j! (n-j)!}.$

Limit ratio test

Theorem 1..5

$\displaystyle \lim_{n \rightarrow \infty} \vert\frac{a_{n+1}}{a_{n}}\vert = r < 1 \Longrightarrow \lim_{n \rightarrow \infty} a_{n} = 0$

NOTE Since $\lim_{n \rightarrow \infty} \vert\frac{a_{n+1}}{a_{n}}\vert = r$ , for all $n$ such that $n \geq N$ , we have

$\displaystyle \vert a_{N+1}\vert$	$\displaystyle \leq$	$\displaystyle r\vert a_{N}\vert$
$\displaystyle \vert a_{N+2}\vert$	$\displaystyle \leq$	$\displaystyle r\vert a_{N+1}\vert \leq r^2\vert a_{N}\vert$
	$\displaystyle \vdots$
$\displaystyle \vert a_{N+n}\vert$	$\displaystyle \leq$	$\displaystyle r\vert a_{N+n-1}\vert \leq r^{n}\vert a_{N}\vert$

Noting that $0 \leq r < 1$ , we have $\lim_{n \to \infty} r^{n} = 0$ . Thus, $\lim_{n \rightarrow \infty} a_{n} = 0.$

Limit with absolute value
$r = \lim{n \to \infty}\vert\frac{a_{n+1}}{a_{n}}\vert$ implies $r \geq 0$ .

Example 1..20 Find the limit of the sequence $a_{n+1} = \sqrt{3a_{n}}, a_{1} = 1$

SOLUTION Note that if $\{a_{n}\}$ converges to $\alpha$ , then $\{a_{n+1}\}$ converges to $\alpha$ . Thus we have

$\displaystyle \alpha = \sqrt{3\alpha}$

Solve this to get $\alpha = 0, 3$ . Since $a_{1} = 1$ , it is impossible to have $\alpha = 0$ . So we have $\alpha = 3$ . Note this is not the end of proof. We have to show $\lim_{n \to \infty}a_{n} = 3$ .

Let $b_{n} = a_{n} -3$ . If we can show $\lim_{n \to \infty}\vert\frac{b_{n+1}}{b_{n}}\vert < 1$ , then we can show $\lim_{n \to \infty}a_{n} = 3$ .

By the limit ratio test, it is enough to show $\lim_{n \to \infty}\vert\frac{a_{n+1} - 3}{a_{n} - 3}\vert < 1$ .

$\displaystyle \lim_{n \to \infty}\vert\frac{a_{n+1} - 3}{a_{n} - 3}\vert$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\vert\frac{\sqrt{3a_{n}} + 3}{a_{n} - 3}\vert = \lim_{n \to \infty}\vert\frac{3a_{n}- 9}{(\sqrt{3a_{n}} + 3)(a_{n} - 3)}\vert$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\vert\frac{3(a_{n}- 3)}{(\sqrt{3a_{n}} + 3)(a_{n} - 3)}\vert = \lim_{n \to \infty}\vert\frac{3}{\sqrt{3a_{n}} + 3}\vert = 0 < 1.$

Thus, $\lim_{n \to \infty}a_{n} = 3$ $\blacksquare$

Exercise 1..20 Find the limit the following.

$\displaystyle \lim_{n \to \infty} \frac{2^{n}}{n!}$

Limit of factorial To find the limit invloving factorials or power functions, it is useful to use the limit ration theorem.

Check
$\frac{n!}{(n+1)!} = \frac{n!}{(n+1)n!} = \frac{1}{n+1}$ .

SOLUTION Set $a_{n} = \frac{2^{n}}{n!}$ . Then

$\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{2^{n+1}}{(n+1)!}\cdot \frac{n!}{2^n} = \frac{2}{n+1}$

and

$\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{2}{n+1} = 0.$

Thus by the limit ratio test, $\displaystyle{\lim_{n \to \infty} \frac{2^{n}}{n!} = 0\ensuremath{ \blacksquare}}$

Exercise A

1.

Find the limit of the following sequences:

(a) $\displaystyle{\{a_{n}\} = \{\sqrt{n}\}}$ (b) $\displaystyle{\{a_{n}\} = \{\frac{n+1}{n^{2}}\}}$ (c) $\displaystyle{\{a_{n}\} = \{\frac{n^{2}}{n + 1}\}}$

(d) $\displaystyle{\{a_{n}\} = \{\frac{2^{n} - 1}{2^{n}}\}}$ (e) $\displaystyle{\{a_{n}\} = \{\frac{1}{n} - \frac{1}{n+1}\}}$

2.

Determine the following sequences are bounded or notDetermine also the following sequences are increasing or decreasing.

(a) $\displaystyle{\{a_{n}\} = \{\frac{2}{n}\}}$ (b) $\displaystyle{\{a_{n}\} = \{\sqrt{4 - \frac{1}{n}}\}}$

3.

Find the general term $\{a_{n}\}$ of the following sequences:

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \frac{1}{n+1}a_{n}, n \geq 1}$ (b) $\displaystyle{a_{1} = 1, a_{n+1} = a_{n} + 2, n \geq 1}$

4.

Determine the following sequences converge or not. If it converges, find the limit

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \frac{1}{e}a_{n}, n \geq 1}$ (b) $\displaystyle{a_{1} = 1, a_{n+1} = 2^{n+1}a_{n}}$

Exercise B

1.

Find the limit of the following sequences:

(a) $\displaystyle{\{a_{n}\} = \{n^4 - 3n^3\}}$ (b) $\displaystyle{\{a_{n}\} = \{\frac{3n^{2}+5}{4n^{3} - 1}\}}$ (c) $\displaystyle{\{a_{n}\} = \{\frac{1 - n}{n - \sqrt{n}}\}}$

(d) $\displaystyle{\{a_{n}\} = \{\frac{n(n+2)}{n+1} - \frac{n^{3}}{n^{2}+1}\}}$ (e) $\displaystyle{\{a_{n}\} = \{\sqrt{n+1} - \sqrt{n}\}}$

2.

Show $\displaystyle{\lim_{n \rightarrow \infty}\sqrt[n]{a} = 1}$ for $a > 0$

3.

Using $\displaystyle{\lim_{n \rightarrow \infty}\sqrt[n]{a} = 1}$ , find the limit of the followings:

(a) $\displaystyle{\{ (a^n + b^n)^{\frac{1}{n}}\}}$ for $a > b > 0$ . (b) $\displaystyle{\{a_{n}\} = \{(1+2^{n}+3^{n})^{\frac{1}{n}}\}}$

4.

Determine the following sequences converge or not

(a) $\displaystyle{2,2^{2},2^{3},\cdots,2^{n},\cdots}$ (b) $a_{n}$ is an nth digit approximation of $\sqrt{2}$

2.

Find the limit of the following sequences $\{a_{n}\}$

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \sqrt{3a_{n} + 4}}$ (b) $\displaystyle{a_{1} = 1, a_{2} = 2, a_{n+2} = \sqrt{a_{n+1}a_{n}}}$

3.

Find the limit of the following sequences:

(a) $\displaystyle{a_{n} = (1 - \frac{1}{n^2})^n}$ (b) $\displaystyle{a_n = (1 + \frac{2}{n})^n}$ (c) $\displaystyle{a_n = \frac{2^n}{n!}}$ (Corollary)

(d) $\displaystyle{a_n = \frac{n!}{n^n}}$