Informative line

### Definitions

Learn definition of convergence and divergence, monotonic & limit of a sequence. Practice sequences definition and series calculus.

# Formula for General Term of Sequence

• Sequence is a list of number written in a definite order or having  a particular pattern.
• e.g        2,4,6,8,_ _ _ _

It is usually written as

a1,a2,a3,_ _ _ _ an,_ _ _ _

a1 is called the first term of the sequence

a2 is called the second term of the sequence....

• We will use infinite sequences so each term an will always have a successor an+1.
• Sequence can be defined as a function whose domain is the set of positive integer and an is like $$f(x)$$, in sequences we use the symbol an for the nth term, instead of $$f(x)$$.
• There are three different ways of representing a sequence.

(1) By using the notation $$\{a_n\}^{\infty}_{n = 1}$$

For e.g. the sequence $$\left\{\dfrac{2n}{5n+1}\right\}^{\infty}_{n=1}$$ will mean the sequence $$\left\{\dfrac{2}{6} , \dfrac{4}{11} \dfrac{6}{16},....\dfrac{2n}{5n + 1}...\right\}$$.

(2) By using the defining formula

e.g.  $$a_n= \dfrac{n}{n+3}$$  will mean the sequence  $$\dfrac{1}{4},\dfrac{2}{5}, \dfrac{n}{n+3}...$$

(3) By writing the terms of the sequence

e.g.    $$\dfrac{3}{5}, \dfrac{-4}{25}, \dfrac{5}{125} , \dfrac{-6}{625},\dfrac{7}{3125}....(-1)^{n-1}\dfrac{n+2}{5^n}...$$

#### Write down the first five terms of the sequence $$a_n = \dfrac{2^n}{1+3^{n+1}}$$ .

A $$\dfrac{1}{5}, \dfrac{4}{9}, \dfrac{12}{5}, \dfrac{11}{27},\dfrac{19}{352}$$

B $$\dfrac{1}{2}, \dfrac{4}{7}, \dfrac{15}{11}, \dfrac{19}{23}, \dfrac{521}{32}$$

C $$\dfrac{1}{5}, \dfrac{1}{7}, \dfrac{4}{41}, \dfrac{4}{61}, \dfrac{16}{365}$$

D $$\dfrac{1}{4}, \dfrac{5}{7}, \dfrac{9}{13}, \dfrac{212}{51}, \dfrac{567}{93}$$

×

If $$a_n$$ = some expression in n, just put the values of n in the expression to find the terms.

In this case $$a_n = \dfrac{2^{n}}{1+3^{n+1}}$$

$$\therefore\;\;a_1 = \dfrac{2^1}{1+3^2} = \dfrac{2}{10} = \dfrac{1}{5}$$

$$a_2 = \dfrac{2^2}{1+3^3} = \dfrac{4}{28} = \dfrac{1}{7}$$

$$a_3 = \dfrac{2^3}{1+3^4} = \dfrac{8}{82} = \dfrac{4}{41}$$

$$a_4 = \dfrac{2^4}{1+3^ 5} = \dfrac{16}{244} = \dfrac{4}{61}$$

$$a_5 = \dfrac{2^5}{1+3^6} = \dfrac{32}{1+729} = \dfrac{32}{730} = \dfrac{16}{365}$$

### Write down the first five terms of the sequence $$a_n = \dfrac{2^n}{1+3^{n+1}}$$ .

A

$$\dfrac{1}{5}, \dfrac{4}{9}, \dfrac{12}{5}, \dfrac{11}{27},\dfrac{19}{352}$$

.

B

$$\dfrac{1}{2}, \dfrac{4}{7}, \dfrac{15}{11}, \dfrac{19}{23}, \dfrac{521}{32}$$

C

$$\dfrac{1}{5}, \dfrac{1}{7}, \dfrac{4}{41}, \dfrac{4}{61}, \dfrac{16}{365}$$

D

$$\dfrac{1}{4}, \dfrac{5}{7}, \dfrac{9}{13}, \dfrac{212}{51}, \dfrac{567}{93}$$

Option C is Correct

#### Find a formula for the general term an of the sequence.  $$\dfrac {1}{2}, \dfrac{-4}{3}, \dfrac{9}{4}, \dfrac{-16}{5},\dfrac{25}{6}, .....$$ Assuming that pattern of first few terms continues.

A $$a_n=\dfrac{(-1)^{n+1}\;n^2}{n+1}$$

B $$a_n=\dfrac{(-1)^{n}\;n^2}{n+1}$$

C $$a_n=\dfrac{n+2}{n^2}$$

D $$\dfrac{n^2}{n+2}$$

×

Observe the pattern of numerator and denominator of the term and try to relate them to the term number.

In this case

$$a_1 = \dfrac{1}{2}, a_2=\dfrac{-4}{3} , a_3= \dfrac{9}{4}, a_4= \dfrac{-16}{5}, a_5= \dfrac{25}{6}$$

Observe that the denominator starts with 2 in $$a_1$$, and increases by 1, whereas numerator is a perfect square of consecutive numbers with alternate positive and negative signs.

$$\therefore\;\; a_n = \dfrac{(-1)^{n+1} \; n^2}{n+1}$$

### Find a formula for the general term an of the sequence.  $$\dfrac {1}{2}, \dfrac{-4}{3}, \dfrac{9}{4}, \dfrac{-16}{5},\dfrac{25}{6}, .....$$ Assuming that pattern of first few terms continues.

A

$$a_n=\dfrac{(-1)^{n+1}\;n^2}{n+1}$$

.

B

$$a_n=\dfrac{(-1)^{n}\;n^2}{n+1}$$

C

$$a_n=\dfrac{n+2}{n^2}$$

D

$$\dfrac{n^2}{n+2}$$

Option A is Correct

# Increasing and Decreasing Sequences

• A sequence $$\{a_n\}$$ is called increasing if $$a_n<a_{ n+1}$$ ; for all $$n\geq1$$, that is $$a_1< a_ 2<a_3 .......a_n,$$ where as it is called decreasing if $$a_n > a_{n+1}$$ for all $$n\geq 1$$
• A sequence is said to be monotonic if it is either increasing or decreasing .
• To show that a sequence $$\{a_n\}$$is increasing point $$a_n<a_{n+1} \;\forall \;n$$
• And to show that it is decreasing point
• e.g     $$a_n = \dfrac{5}{n+4}$$ is a decreasing sequence as

$$\dfrac{5}{n+4} > \dfrac{5}{(n+1)+4}\; or \; a_n>a_{n +1}$$

$$\dfrac{5}{n+4} > \dfrac{5}{n+5}$$

#### Which of the following sequence is increasing ?

A $$a_n = \dfrac{1}{5n+4}$$

B $$a_n = (-3)^{n+1}$$

C $$a_n = n +\dfrac{1}{n}$$

D $$a_n= \dfrac{1}{2n+5}$$

×

A sequence is said to be increasing if $$a_n < a_{n+1}$$ for all $$n \geq 1 \; (n \in N)$$.

Check the options

(a)     $$a_n = \dfrac{1}{5n+4}$$

$$a_n < a_{n+1}$$

$$\Rightarrow \; \dfrac{1}{5n+4}\; < \dfrac{1}{5(n+1) +4}$$

or $$\dfrac{1}{5n+4}<\dfrac{1}{5n+9}$$ or $$5n+ 4 > 5n +9$$

or $$4>9$$ which is not true.

$$\therefore \; a_n$$ is a decreasing sequence .

(b)   $$a_n = (-3)^{n+1}$$

$$a_n < a_{n+1}$$

$$\Rightarrow \;(-3) ^{n+1} < (-3) ^{n+2}$$

$$\Rightarrow \; (-1)^{n+1} × 3^{n+1}\;<\; (-1)^{n+2} × 3^{n+2}$$

or  $$(-1)^{n+1} < (-1) ^{n+2} × 3$$

which may or may not be true according to n.

$$\therefore\; a_n$$ is non monotonic .

(c)  $$a_n = n + \dfrac{1}{n}$$

$$a_n <a_{n+1}$$

$$\Rightarrow \; n+ \dfrac{1}{n } < n+1 + \;\dfrac{1}{n+1}$$

or    $$\dfrac{1}{n} < \;1 + \dfrac{1}{n+1} \; or \dfrac{n+1+1}{n+1} - \dfrac{1}{n} > 0$$

$$or \;\; \dfrac{n^2 + 2n -n -1}{n (n+1)} > 0 \;\; or \;\; \dfrac{n^2 + n -1}{n (n+1)} > 0$$

which is true.

$$\therefore\;\; a_n$$ is an increasing sequence.

(d)    $$a_n = \dfrac{1}{2n+5}$$

$$a_n<a_{n+1}$$

$$\Rightarrow \; \dfrac{1}{2n+5}<\dfrac{1}{2 (n+1) +5} \; or \; \dfrac{1}{2n+5} < \dfrac{1}{2n+7}$$

$$or\;\; 2n+5 > 2n+7\;\; or\;\; 5>7$$  which is not true.

$$\therefore \; \; a_n$$ is a decreasing sequence.

### Which of the following sequence is increasing ?

A

$$a_n = \dfrac{1}{5n+4}$$

.

B

$$a_n = (-3)^{n+1}$$

C

$$a_n = n +\dfrac{1}{n}$$

D

$$a_n= \dfrac{1}{2n+5}$$

Option C is Correct

# Finding Limit for Convergent and Divergent Sequence

A sequence $$\{a_n\}$$ has the limit L and we write $$\lim\limits_{n\to \infty \; }a_n = L \;\;\text{or}\;\;a_n \to L$$ as $$n \to \infty$$ if we

can make the term an as close to L as we like by taking n sufficiently large.

• If $$\lim\limits _{n\to \infty }$$ an exists or it is a finite number we say that sequence converges ( or is convergent ), otherwise we say the sequence diverges ( or is divergent ).

Pattern of a converging sequence when $$\lim\limits_{n\to\infty}\;\;a_n = L$$  Pattern of a divergent sequence when $$\lim\limits_{n \to \infty }\; a_n$$ does not exist.

•  If $$\lim \limits _{n\to \infty }\; f (n) = L$$ and $$f (n) = a_n$$ when n is an integer, then $$\lim\limits _{n \to \infty } \; a_n = L$$.  #### Find the limit of the following convergent sequence.  $$a_n\; = \sqrt{\dfrac{5n+3}{2n+5}}$$

A $$L = \sqrt {\dfrac{5}{2}}$$

B $$L = \dfrac{5}{2}$$

C $$L= \dfrac{1}{2}$$

D $$L = \sqrt {\dfrac{5}{3}}$$

×

For a convergent sequence

$$\lim\limits _{n\to \infty} \; a_n\; = L$$    where   L is a finite number, called the limit of the sequence .

In this case

$$L = \lim\limits _{n \to \infty } \sqrt{\dfrac{5n + 3}{2n + 5}}$$

$$= \lim\limits _{n\to \infty} \sqrt {\dfrac{5 + \dfrac{3}{n}}{2 + \dfrac{5}{n}}}$$  ( Divide Numerator & Denominator by n in the square root )

$$= \sqrt {\dfrac{5+0}{2+0}} \;\;\;=\;\;\;\sqrt{\dfrac{5}{2}}$$

$$\therefore \;\;L = \sqrt {\dfrac{5}{2}}$$

### Find the limit of the following convergent sequence.  $$a_n\; = \sqrt{\dfrac{5n+3}{2n+5}}$$

A

$$L = \sqrt {\dfrac{5}{2}}$$

.

B

$$L = \dfrac{5}{2}$$

C

$$L= \dfrac{1}{2}$$

D

$$L = \sqrt {\dfrac{5}{3}}$$

Option A is Correct

#### Find the limit of the following convergent sequence.  $$a_n = ln(3n^2 + 1)\;- ln \; (n^2 + 4)$$

A L = ln 2

B L = ln 3

C L = ln 5

D L = ln 6

×

For a convergent sequence

$$\lim\limits _{n \to \infty} \; a_n\; = L$$    where   L is a finite number, called the limit of the sequence.

In this case

$$a_n = ln (3n^2 +1) - ln (n^2 +4)$$

$$= ln \left(\dfrac{3n^2 + 1}{n^2 +4}\right)$$        ( use ln a – ln b = $$ln\dfrac{a}{b}$$ )

$$\therefore$$ If $$a_n = f (n)$$ then $$f(n) =ln \left(\dfrac{3n^2 + 1}{n^2 + 4}\right)$$

$$\therefore \; L = \lim\limits _{n\to \infty }\; ln \left(\dfrac{3n^2 + 1}{n^2 + 4}\right)$$

$$= \lim\limits _{n\to\infty } \; ln \left(\dfrac{3+\dfrac{1}{n^2}}{1+\dfrac{4}{n^2}}\right)\;$$

$$= ln \;\dfrac{3+0}{1+0}\\ = ln\; 3$$

### Find the limit of the following convergent sequence.  $$a_n = ln(3n^2 + 1)\;- ln \; (n^2 + 4)$$

A

L = ln 2

.

B

L = ln 3

C

L = ln 5

D

L = ln 6

Option B is Correct