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

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 . 

 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 + a_{2 +} a₃

S_n = a₁ + a_{2 +} a₃ + ....a_n = \(\sum\limits ^n_{i=1} a_i\)

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

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

• Given a series \(\sum \limits _1^\infty\; a_n = a_1 +a_2 +a_3 +........a_n +.....\)
•  Let S_n  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  

• 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

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

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

• 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.

• 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.