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 \(\sum \; a_n\)

Illustration Questions

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  

Illustration Questions

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

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.

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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

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

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

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

Illustration Questions

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

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. 

Illustration Questions

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

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