Informative line

### Series

Learn geometric, telescoping convergence & divergence of a Series. Practice an important example of infinite series is the geometric series.

# Series ( Definition )

• If we add the terms of an infinite sequence $$\{a_n\}^\infty_{n=1}$$ we get an expression                                                                                                                                                                                           a1 + a2 + a3 .......... a+ .............

Which is called an infinite series and is derived in short by

$$\sum\limits^\infty_{n=1} \;\;a_n\;\; or \;\; \sum \; a_n$$

•  Some infinite series will have finite sum, i.e we add more and more terms the sum gets closer to a quantity but never reaches that quantity.

$$e.g. \;\; \dfrac{1}{2} +\dfrac{1}{4} +\dfrac{1}{6} + ........\dfrac{1}{2^n} + ...... = \sum\limits^\infty_1 \; \dfrac{1}{2^n} = 1$$

We say that sum of this infinite series is 1.The following table shows the sum of series for a particular no. of term n .

n Sum
1 .5
2 .75
3 .875
4 .9375
5 .96875
6 .984375
7 .99218750
10 .99902344
15 .99996948
20 .99999905
25 .99999997

Some infinite series will not have a finite sum.

e.g.     1+2+3+ .........n+.......= $$\sum\limits^\infty_{1} \;n$$

The sum to this series is $$S_n$$$$\dfrac {n(n+1)}{2}$$

The consider partial sum of the series

S1 = a1

S2 = a1 + a2

S3 = a1 + a2 + a3

Sn = a1 + a2 + a3 + ....an = $$\sum\limits ^n_{i=1} a_i$$

These partial sum form a new sequence which may or may not have a limit.

If $$\lim\limits _{n\to\infty}$$ Sn = S ( a finite number ) we call it the sum of infinite series $$\sum \; a_n$$

#### Calculate the fifth term of sequence of partial sum for the following series current to two decimal places. $$\sum\limits^\infty_{n=1}\; \dfrac{n^2}{2+\sqrt n}$$

A $$S_5 = 42.62$$

B $$S_5 = 1.29$$

C $$S_5 = 13 .81$$

D $$S_5 = 16 .72$$

×

Partial sum = Sn = a1 + a2 +..........an

In this case the fifth partial sum

$$=\; S_5 = a_1+a_2+a_3+a_4+a_5$$

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

$$\therefore\;\; S_5 = \dfrac{1^2}{2+\sqrt 1} + \dfrac{2^2}{2 +\sqrt2}+\dfrac{3^2}{2+\sqrt3} + \dfrac{4^2}{2+\sqrt 4} +\dfrac{5^2}{2+\sqrt 5}$$

$$=\dfrac{1}{3} + \dfrac{4}{3.41}+ \dfrac{9}{3.73}+ \dfrac{16}{4}+ \dfrac{25}{4.24}$$

$$= .33 +1.17 +2.41+4+5.90 =13.81$$

$$\therefore \;\; S_5 = 13.81$$

### Calculate the fifth term of sequence of partial sum for the following series current to two decimal places. $$\sum\limits^\infty_{n=1}\; \dfrac{n^2}{2+\sqrt n}$$

A

$$S_5 = 42.62$$

.

B

$$S_5 = 1.29$$

C

$$S_5 = 13 .81$$

D

$$S_5 = 16 .72$$

Option C is Correct

# Convergence and Divergence of a Series

• Given a series $$\sum \limits _1^\infty\; a_n = a_1 +a_2 +a_3 +........a_n +.....$$
•  Let Sn  denote it partial sum.

$$S_n = \sum \limits _{i=1}^n a_i = a_1 +a_2 +.....a_n$$

• If the sequence $$\{S_n\}$$ is convergent and $$\lim\limits_{n\to \infty}\; S_n = S$$ exists as a finite real number, then we say that series $$\sum a_n$$ is convergent and we write

$$a_1+a_2+......a_n+....= S\;\;or\;\; \sum \limits^\infty_{n-1} \; a_n = S$$

The number $$S$$ is called the sum of the series.

• If the sequence $$\{S_n\}$$is divergent, then we say that series is divergent

#### Calculate the sum of the series whose partial sum is given by $$S_n = \dfrac{5n^2 +3}{2n^2-1}$$

A $$S= \dfrac{2}{5}$$

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

C $$S= \dfrac{15}{4}$$

D $$S = \dfrac{-1}{6}$$

×

Sum of series $$= S= \lim\limits _{n\to\infty}\; S_n$$

In this case , $$S_n= \dfrac{5n^2 +3 }{2n^2-1}$$

$$\therefore\; S = \lim\limits_{n\to \infty } \dfrac{5n^2 +3}{2n^2 -1}\; = \lim \limits _{n\to\infty} \dfrac{5 + \dfrac{3}{n^2}}{2-\dfrac{1}{n^2}}$$

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

$$\therefore \;\; S = \dfrac{5}{2}$$

### Calculate the sum of the series whose partial sum is given by $$S_n = \dfrac{5n^2 +3}{2n^2-1}$$

A

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

.

B

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

C

$$S= \dfrac{15}{4}$$

D

$$S = \dfrac{-1}{6}$$

Option B is Correct

#### Determine whether the given series is convergent or divergent, find the sum if it is convergent.  $$\sum\limits _{n=1}^\infty\;\; \left(\dfrac{2+3^n}{5^n}\right)$$

A 2

B 5

C 4

D $$\dfrac{-3}{2}$$

×

The sum of geometric series

$$a+ar +ar^2+...... ar^{n-1} +..... \sum \limits^0_{n=1}\; ar^{n-1}$$

$$= \dfrac{a}{1-r} \; \;\;\;if \; \;\;\;|r| < 1$$

Observe that  $$\sum \limits ^\infty_{1} \; \dfrac{2+3^n}{5^n}$$  is a geometric series

$$\sum \limits ^\infty_{n=1} \; \left(\dfrac{2+3^n}{5^n}\right)= \sum \limits ^\infty_{1}\dfrac{2}{5^n} + \sum \limits ^\infty_1\left(\dfrac{3}{5}\right)^n$$

$$\underbrace{= 2 \left(\sum \limits^\infty_1\left(\dfrac{1}{5}\right)^n\right)}_\text{a= 1/5, r =1/5} \;+\;\;\underbrace{ \left(\sum \limits ^\infty_1 \left(\dfrac{3}{5}\right)^n \right)}_\text{a= 3/5, r =3/5}$$

Both geometric series are convergent as  $$|r| < 1$$.

$$\therefore \;\; S = \lim\limits _{n\to \infty }\; S_n = \dfrac{a}{1-r}$$

Sum of 1st G.P. $$= \dfrac{1/5}{1-1/5} = \dfrac{1}{4}$$

Sum of 2st G.P. $$= \dfrac{3/5}{1-3/5} = \dfrac{3}{2}$$

$$\therefore\;\;$$Required sum $$= 2 × \dfrac{1}{4} + \dfrac{3}{2}= 2$$

### Determine whether the given series is convergent or divergent, find the sum if it is convergent.  $$\sum\limits _{n=1}^\infty\;\; \left(\dfrac{2+3^n}{5^n}\right)$$

A

2

.

B

5

C

4

D

$$\dfrac{-3}{2}$$

Option A is Correct

# Expressing a Recalling Decimal as Ratio of Integers

• Suppose we have a number

$$5.\overline{612} = 5.612612612.....$$

• Let  $$x = 5. \overline {612}$$ then $$1000 x = 5612 .\overline {612} \;\;\;\;— (1)$$

Subtracting equation (1) form (2) we get

$$999x = 5612.\overline {612}-5. \overline {612}$$

$$\Rightarrow \;\;999x = 5607$$

$$\Rightarrow \;\;x = \dfrac{5607}{999}$$

• In this way we can express any recurring decimal in the form of $$\dfrac{p}{q}$$ where p,q are integers.
• We can also say that

$$x = 5. \overline {612} = 5. 612612612..... = 5 + \dfrac{612}{10 ^3}+ \dfrac{612}{10^6}+.....$$

$$= 5 + \dfrac{612/10^3}{1- 1/10^3} = 5 +\dfrac{612}{999} = \dfrac{5607}{999}$$

#### Express the following numbers as ratio of integers. $$3.\overline {423}$$

A $$\dfrac{5128}{999}$$

B $$\dfrac{3420}{999}$$

C $$\dfrac{3428}{999}$$

D $$\dfrac{512}{999}$$

×

Write the number as an appropriate geometric series and then apply the sum of geometric series formula.

$$3.\overline {423} = 3.423 423423.....$$

$$= 3+ \dfrac{423}{10 ^3} + \dfrac{423}{10 ^6} +\dfrac{423}{10^9}+ ....$$

$$=3+ \text{(geometric series with a }= \dfrac{423}{10^3} \text{ and } r=\dfrac{1}{10^3})$$

$$= 3 +\dfrac{\dfrac{423}{10^3}}{1-\dfrac{1}{10^3}}$$

$$= 3 + \dfrac{423}{999} = \dfrac{2997 + 423}{999} = \dfrac{3420}{999}$$

### Express the following numbers as ratio of integers. $$3.\overline {423}$$

A

$$\dfrac{5128}{999}$$

.

B

$$\dfrac{3420}{999}$$

C

$$\dfrac{3428}{999}$$

D

$$\dfrac{512}{999}$$

Option B is Correct

# Geometric Series

• An important example of infinite series is the geometric series

$$a+ar+ar^2+ .......ar^{n-1} + .......=\sum \limits ^\infty_{n=1}\; ar^{n-1} (a\neq 0)$$

Note that each term is obtained by multiplying note preceding tern by r called common ratio fof this G.P.

• Consider $$S_n = a+ ar+ar^2 + .....ar^{n-1}\;\;\;\;\; — (1)$$

$$rS_n = ar + ar^2 + .... ar^{n=1} + ar^n \;\;\;\;\;\; — (2)$$

Equation  (1)  –  Equation  (2)  $$\Rightarrow \; S_n ( 1-r ) = a- ar ^n$$

$$\Rightarrow \; S_n = \dfrac{a(1-r^n)}{1-r}$$

Now $$r^n\rightarrow 0 \;\;as\;\; n \rightarrow \infty \; \text{only if }\; -1<r < 1$$

$$\therefore$$ For  $$-1<r<1 , \; \lim\limits _{n\to \infty} \; S_n = \dfrac{a}{1-r}$$

$$\therefore\;\; \sum \limits ^\infty_{n=1}\; ar^{n-1} \;\; = \; a+ar+ar^2 +..... ar^{n-1} .....$$

Is convergent if $$|r|<1\; or - 1 < r < 1$$ and its sum is given by

$$\sum \limits ^\infty _{n = 1}\; ar ^{n-1}\; = S = \dfrac{a}{1-r}$$

• If $$|r| \geq 1$$ then the geometric series diverges.  #### Find the sum of the following convergent geometric series.  $$4 +3+\dfrac{9}{4} + \dfrac{27}{16} + ......$$

A $$S = 4$$

B $$S= 16$$

C $$S = \dfrac{5}{2}$$

D $$S = \dfrac{19}{2}$$

×

The sum of geometric series

$$a+ar +ar^2+...... ar^{n-1} +..... \sum \limits^0_{n=1}\; ar^{n-1}$$

$$= \dfrac{a}{1-r} \; \;\;\;if \; \;\;\;|r| < 1$$

In this case series is

$$4+3+\dfrac{9}{4} + \dfrac{27}{16} +.....$$

Comparing with $$a+ar+ar^2 +.....$$

We get $$a=4 , \; \;\;r= \dfrac{3}{4}$$

$$\therefore \;\; S = \lim\limits _{n\to \infty }\; S_n = \dfrac{a}{1-r} = \dfrac{4}{1-\dfrac{3}{4}}\;\; =16$$

$$\therefore \;\; S =16$$

### Find the sum of the following convergent geometric series.  $$4 +3+\dfrac{9}{4} + \dfrac{27}{16} + ......$$

A

$$S = 4$$

.

B

$$S= 16$$

C

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

D

$$S = \dfrac{19}{2}$$

Option B is Correct

#### For what values of $$x$$ the following series will converge. $$\sum\limits _{n=0}^\infty\;\; \dfrac{(x-3)^n}{4^n}$$

A $$x\in(-2,6)$$

B $$x\in (-1,7)$$

C $$x\in \left(\dfrac{1}{2}4\right)$$

D $$x\in (5, \infty)$$

×

A geometric series  $$\sum\limits ^\infty_{n=1}\; ar^{n-1} \;\; or \;\;\sum\limits^\infty _{n=0} \; ar^n$$

Will converge if  $$|r|<1\;\; or \;\; -1 < r < 1$$

In this case the given series $$\sum \limits ^\infty_{n=0} \; \dfrac{(x-3)^n}{4^n}$$

$$= \; \sum\limits _{n=0}^\infty\;\; \left(\dfrac{x-3}{4}\right)^n$$ is a geometric series with

$$a = 1 , \;\;\;r = \dfrac{x-3}{4}$$

The geometric series will converge if $$|r| < 1 \;\; or \;\; - 1 < r < 1$$

$$\Rightarrow \;\; -1 < \dfrac{x-3}{4} < 1 \;\;\; or \;\;\; -4 < x -3 < 4$$

.$$or \; -1 < x < 7\;\;$$

$$\Rightarrow \;\; x\in(-1,7)$$

### For what values of $$x$$ the following series will converge. $$\sum\limits _{n=0}^\infty\;\; \dfrac{(x-3)^n}{4^n}$$

A

$$x\in(-2,6)$$

.

B

$$x\in (-1,7)$$

C

$$x\in \left(\dfrac{1}{2}4\right)$$

D

$$x\in (5, \infty)$$

Option B is Correct

# Convergence of a Series Using Telescopic Sum

• Consider the series $$\sum \limits_{n=1}^{\infty}\; \dfrac{1}{n (n+1)}$$
• We write this series as

$$\sum \limits _{n=1}^\infty \; \dfrac{1}{n(n+1)} = \sum\limits _1^\infty \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right)$$

.$$\left(\dfrac{1}{1} -\dfrac{1}{2}\right)+\left(\dfrac{1}{2} -\dfrac{1}{3}\right)+....\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+....$$

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

As $$n\to\infty$$ this will tends to 1 as $$\dfrac{1}{n+1}\rightarrow0$$ .

• So we say that the above series converges and sum of series is 1. The sum in which terms of the series cancel in pairs is called telescopic sum.

#### Find the sum of the following convergent series by expressing Sn as telescopic sum. $$\sum\limits_{n=1}^\infty \; \left(2^{1/n}-2 ^{1/n+1}\right)$$

A 1

B 2

C –2

D 5

×

Write each term of the series as difference of two terms such that terms cancel out in pairs and we are left with only two terms.

$$\sum\limits_{n=1}^\infty \; \left(2^{1/n}-2 ^{1/n+1}\right) = \left(2^1-2^{1/2}\right)+ \left(2^{1/2}-2 ^{1/3}\right) +\left(2^{1/3} - 2 ^{1/4}\right) + ....\left(2^{1/n} - 2^ {1/n +1}\right)+....$$

$$\Rightarrow \;\; S_n = \left(2^1-2^{1/2}\right)+\left(2^{1/2}-2 ^{1/3}\right)+\left(2^{1/3}-2 ^{1/4}\right)+....\left(2^{1/n}-2 ^{1/n +1}\right)$$

$$= 2-2 ^{1/n+1}$$

$$\therefore\;\; S = \lim\limits _{n\to \infty}\; S_n = \lim\limits _{n\to\infty}\;\; 2-2 ^{1/n+1}$$

$$= 2 -\lim\limits_{n \to \infty} 2^{1/n+1}\;=2 - 2^0 = 2 -1 = 1$$

### Find the sum of the following convergent series by expressing Sn as telescopic sum. $$\sum\limits_{n=1}^\infty \; \left(2^{1/n}-2 ^{1/n+1}\right)$$

A

1

.

B

2

C

–2

D

5

Option A is Correct