In the previous section, we saw infinite series related to binomial expansions. In this section, we will see infinite series related to geometric series.
• In chapter 9, we saw the general form of a geometric series. Let us write it again:
a+ar+ar2+ar3 + . . . + arn−1
• This is a finite series. There are n terms.
• We know that, for such a finite series, the sum is given by:
Sn=a(1−rn)1−r
• Now let us suppose that, there are infinite number of terms. Then it will be an infinite geometric series.
• For such series also, we can derive a formula to find the sum.
• First we will see an example to learn the general features of infinite geometric series. It can be written in 7 steps:
1. Consider the geometric series:
1+23+49+ . . .
2. We see that, the first term, a = 1
3. Let us calculate r:
• 23÷1=23
• 49÷23=49×32=23
• So we can write: r=23
4. Now we can write the sum of n terms:
Sn = a(1−rn)1−r = 1×[1−(23)n]1−(23) = [1−(23)n]13 = 3[1−(23)n]
5. We know that, when the number of terms in the series increases, n increases.
• When n increases, what happens to the term (23)n ?
◼ Let us put some convenient values for n:
(calculations can be easily done using a spreadsheet program)
• When n = 1, we get:
(23)n=(23)1=0.666666666666667
• When n = 5, we get:
(23)n=(23)5=0.131687242798354
• When n = 10, we get:
(23)n=(23)10=0.017341529915833
• When n = 20, we get:
(23)n=(23)20=0.000300728659822
• When n = 30, we get:
(23)n=(23)30=0.00000521509505084655
6. It is clear that, when n increases, (23)n decreases.
• When n approaches infinity, (23)n will approach zero. There will be more than a million zeros just to the right of the decimal point. For all practical purposes, it can be considered as zero.
7. Now consider the result in (4).
• Using that result, we can write:
Sn=3[1−0]=3 where n is infinity
Let us see if we can write a general form of the above result. It can be done in 4 steps:
1. Consider the geometric series:
a+ar+ar2+ar3 + . . . + arn−1 + . . .
• The common ratio r is a proper fraction. It can be +ve or -ve.
• Mathematically, we write this as: |r| < 1
(Proper fraction is a fraction in which denominator is larger than the numerator)
• We want the sum when n is infinity.
2. We have:
Sn=a(1−rn)1−r
• This can be written as:
Sn=a1−r − arn1−r
3. Now we note an important point:
♦ Since |r| < 1,
♦ rn will approach zero
♦ when n approach infinity.
• So the result in (2) will become:
Sn = a1−r − arn1−r = a1−r − a×01−r = a1−r − 0 = a1−r
4. Symbolically, when sum of an infinite geometric series is written, we use "S" instead of "Sn".
• So we have: S=a1−r
Now we will see a solved example:
Solved example A.2
Find the sum to infinity of the following geometric progressions:
(i) 1,12,122,123, . . .
(ii) 1,−12,122,−123, . . .
(iii) −54,516,−564, . . .
Solution:
Part (i):
1. The infinite geometric series associated with the given G.P is:
1+12+122+123+ . . .
2. First term a is 1 and the common ratio r is 12.
3. Absolute value of r is less than 1. So we can use the formula: S=a1−r
S = a1−r = 11−12 = 2
Part (ii):
1. The infinite geometric series associated with the given G.P is:
1−12+122−123+ . . .
2. First term a is 1 and the common ratio r is −12.
3. Absolute value of r is less than 1. So we can use the formula: S=a1−r
S = a1−r = 11−(−12) = 11+12 = 132 = 23
Part (iii):
1. The infinite geometric series associated with the given G.P is:
−54+516−564+ . . .
2. First term a is −54 and the common ratio r is −14.
3. Absolute value of r is less than 1. So we can use the formula: S=a1−r
S = a1−r = −541−(−14) = −541+14 = −5454 = −1
In the next section, we will see exponential series.
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