Learn geometric, telescoping convergence & divergence of a Series. Practice an important example of infinite series is the geometric series.
Which is called an infinite series and is derived in short by
\(\sum\limits^\infty_{n=1} \;\;a_n\;\; or \;\; \sum \; a_n \).
\(e.g. \;\; \dfrac{1}{2} +\dfrac{1}{4} +\dfrac{1}{6} + ........\dfrac{1}{2^n} + ...... = \sum\limits^\infty_1 \; \dfrac{1}{2^n} = 1\)
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 \(\sum \; a_n\).
A \(S_5 = 42.62\)
B \(S_5 = 1.29\)
C \(S_5 = 13 .81\)
D \(S_5 = 16 .72\)
\(S_n = \sum \limits _{i=1}^n a_i = a_1 +a_2 +.....a_n\).
\(a_1+a_2+......a_n+....= S\;\;or\;\; \sum \limits^\infty_{n-1} \; a_n = S\)
The number \(S\) is called the sum of the series.
A \(S= \dfrac{2}{5}\)
B \(S= \dfrac{5}{2}\)
C \(S= \dfrac{15}{4}\)
D \(S = \dfrac{-1}{6}\)
\(5.\overline{612} = 5.612612612.....\)
Subtracting equation (1) form (2) we get
\(999x = 5612.\overline {612}-5. \overline {612}\)
\(\Rightarrow \;\;999x = 5607\)
\(\Rightarrow \;\;x = \dfrac{5607}{999}\)
\(x = 5. \overline {612} = 5. 612612612..... = 5 + \dfrac{612}{10 ^3}+ \dfrac{612}{10^6}+.....\)
\(= 5 + \dfrac{612/10^3}{1- 1/10^3} = 5 +\dfrac{612}{999} = \dfrac{5607}{999}\)
A \(\dfrac{5128}{999}\)
B \(\dfrac{3420}{999}\)
C \(\dfrac{3428}{999}\)
D \(\dfrac{512}{999}\)
\(a+ar+ar^2+ .......ar^{n-1} + .......=\sum \limits ^\infty_{n=1}\; ar^{n-1} (a\neq 0)\)
Note that each term is obtained by multiplying note preceding tern by r called common ratio fof this G.P.
\(rS_n = ar + ar^2 + .... ar^{n=1} + ar^n \;\;\;\;\;\; — (2)\)
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
\(\sum \limits ^\infty _{n = 1}\; ar ^{n-1}\; = S = \dfrac{a}{1-r}\)
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 \(x\in (5, \infty)\)
\(\sum \limits _{n=1}^\infty \; \dfrac{1}{n(n+1)} = \sum\limits _1^\infty \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right)\)
.\(\left(\dfrac{1}{1} -\dfrac{1}{2}\right)+\left(\dfrac{1}{2} -\dfrac{1}{3}\right)+....\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+....\)
\(= 1- \dfrac{1}{n+1}\)
As \(n\to\infty\) this will tends to 1 as \(\dfrac{1}{n+1}\rightarrow0\) .