Limit of Sequence

Sequences

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

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

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.

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.

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

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

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

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.

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

SOLUTION Since $a_{1} = 1, a_{2} = \sqrt{3a_{1}} = \sqrt{3}$ , we have $a_{1} < a_{2}$. Now use mathematical induction. ^1.5

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

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

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

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

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.

Limit properties

Theorem 1..1   Suppose that $${\lim_{n \rightarrow \infty}a_{n} = a, \lim_{n \rightarrow \infty}b_{n} = b}$$ and $c$ is constant. Then we have the followings:

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

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

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

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

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

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

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

Limit properties

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

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

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

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.

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

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

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

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

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

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

SOLUTION

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus by the squeezing theorem,

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

Note that

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

we have

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

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

SOLUTION 1. $${\lim_{n \to \infty}\frac{(-2)^n}{1 + 3^n} = \lim_{n \to \infty}\f... ...rac{-2}{3}\big)^n}{\big(\frac{1}{3}\big)^n + 1} = 0\ensuremath{\ \blacksquare}}$$ 2. $${\lim_{n \to \infty}\frac{5^{n+1} + 4^n}{3^{n-1} - 5^n} = \lim_{n ... ...big(\frac{3}{5}\big)^{n-1} - 1} = \frac{5}{-1} = -5\ensuremath{\ \blacksquare}}$$

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

Bernoulli inequality

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

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

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,

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

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

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

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

Now for $r = 1$, $a_{n} = \frac{1}{2}$ and $${\lim_{n \to \infty}a _{n} = \frac{1}{2}}$$. 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 $r > 1, r < -1$,

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

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

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

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

Then,

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

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

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

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

Monotone convergence theorem

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

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 $${a_{n+1} = \sqrt{3a_{n}}, a_{1} = 1}$$. Determine the sequence $\{a_{n}\}$ converges or diverges..

SOLUTION We show by induction that the sequence $\{a_{n}\}$ converges. 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 $${a_{n} = (1 + \frac{1}{n})^n}$$. Determine the sequence $\{a_{n}\}$ converges or diverges.

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

Expand $${(1 + \frac{1}{n})^n}$$ ^1.6using binomial theorem,

$$a_{n} = \left(1 + \frac{1}{n}\right)^{n} = \binom{n}{0} \left(\frac{1}{n}... ...binom{n}{1} \left(\frac{1}{n} \right) + \binom{n}{2} \left(\frac{1}{n}\right)^2$$

$$+ \cdots + \binom{n}{n} \left(\frac{1}{n}\right)^n$$

$$= 1 + n\cdot \frac{1}{n} + \frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)^n$$

$$< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots < 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \leq 3.$$

Thus, the sequence $\{a_{n}\}$ is bounded above increasing sequnce and converges $\ \blacksquare$

Euler e

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

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

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

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

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

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

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

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

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

$$= \frac{1}{e}\ensuremath{\ \blacksquare}$$

Limit ratio test

Theorem 1..5  

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

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

$$\vert a_{N+1}\vert \leq r\vert a_{N}\vert$$

$$\vert a_{N+2}\vert \leq r\vert a_{N+1}\vert \leq r^2\vert a_{N}\vert$$

$$\vdots$$

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

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

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

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

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

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

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

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

Exercise 1..20   Find the limit the following.

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

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

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

and

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

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

Exercise A

1.

Find the limit of the following sequences:?D

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

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

2.

Determine the following sequences are bounded or not?DDetermine also the following sequences are increasing or decreasing.

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

3.

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

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

(c) $${a_{1} = 1, a_{n+1} = a_{n} + 2n + 1, \ n \geq 1}$$

4.

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

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

(c) $${a_{1} = 1, a_{n+1} = \frac{n}{n+1}a_{n}}$$

Exercise B

1.

Find the limit of the following sequences:

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

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

2.

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

3.

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

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

4.

Determine the following sequences converge or not?D

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

2.

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

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

3.

Find the limit of the following sequences:

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

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