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

- If we add the terms of an infinite sequence \(\{a_n\}^\infty_{n=1}\) we get an expression a
_{1}+ a_{2}+ a_{3}.......... 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 .

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

S_{1} = a_{1}

S_{2} = a_{1} + a_{2}

S_{3} = a_{1} + a_{2 + }a_{3}

S_{n} = a_{1} + a_{2 + }a_{3} + ....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 \(\sum \; a_n\).

A \(S_5 = 42.62\)

B \(S_5 = 1.29\)

C \(S_5 = 13 .81\)

D \(S_5 = 16 .72\)

- 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

A \(S= \dfrac{2}{5}\)

B \(S= \dfrac{5}{2}\)

C \(S= \dfrac{15}{4}\)

D \(S = \dfrac{-1}{6}\)

- 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

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

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

A \(\dfrac{5128}{999}\)

B \(\dfrac{3420}{999}\)

C \(\dfrac{3428}{999}\)

D \(\dfrac{512}{999}\)

- 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.

A \(S = 4\)

B \(S= 16\)

C \(S = \dfrac{5}{2}\)

D \(S = \dfrac{19}{2}\)

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

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

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

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

- 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.