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:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}$
• This is a finite series. There are n terms.
• We know that, for such a finite series, the sum is given by:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$
• 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:
$1 + \frac{2}{3} + \frac{4}{9} + ~.~.~.$
2. We see that, the first term, a = 1
3. Let us calculate r:
• $\frac{2}{3} \div 1 = \frac{2}{3}$
• $\frac{4}{9} \div \frac{2}{3} = \frac{4}{9} \times\frac{3}{2} = \frac{2}{3}$
• So we can write: $r = \frac{2}{3}$
4. Now we can write the sum of n terms:
$\begin{array}{ll}{} &{S_n} & {~=~} &{\frac{a \left(1 - r^n \right)}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{1 \times \left[1 - \left(\frac{2}{3} \right)^n \right]}{1-\left(\frac{2}{3} \right)}} &{} \\
{} &{} & {~=~} &{\frac{\left[1 - \left(\frac{2}{3} \right)^n \right]}{\frac{1}{3}}} &{} \\
{} &{} & {~=~} &{3 \left[1 - \left(\frac{2}{3} \right)^n \right]} &{} \\
\end{array}$
5. We know that, when the number of terms in the series increases, n increases.
• When n increases, what happens to the term $\left(\frac{2}{3} \right)^n$ ?
◼ Let us put some convenient values for n:
(calculations can be easily done using a spreadsheet program)
• When n = 1, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^1 = 0.666666666666667$
• When n = 5, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^5 = 0.131687242798354$
• When n = 10, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{10} = 0.017341529915833$
• When n = 20, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{20} = 0.000300728659822$
• When n = 30, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{30} = 0.00000521509505084655$
6. It is clear that, when n increases, $\left(\frac{2}{3} \right)^n$ decreases.
• When n approaches infinity, $\left(\frac{2}{3} \right)^n$ 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:
$S_n = 3[1 - 0] = 3~\text{where n is infinity}$
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:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}~+ ~.~.~.$
• 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:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$
• This can be written as:
$S_n = \frac{a}{1-r}~-~\frac{a r^n}{1-r}$
3. Now we note an important point:
♦ Since |r| < 1,
♦ $r^n$ will approach zero
♦ when n approach infinity.
• So the result in (2) will become:
$\begin{array}{ll}{} &{S_n} & {~=~} &{\frac{a}{1-r}~-~\frac{a r^n}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{a}{1-r}~-~\frac{a \times 0}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{a}{1-r}~-~0} &{} \\
{} &{} & {~=~} &{\frac{a}{1-r}} &{} \\
\end{array}$
4. Symbolically, when sum of an infinite geometric series is written, we use "S" instead of "Sn".
• So we have: $S = \frac{a}{1-r}$
Now we will see a solved example:
Solved example A.2
Find the sum to infinity of the following geometric progressions:
(i) $1, \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3},~.~.~.$
(ii) $1, \frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3}, ~.~.~.$
(iii) $\frac{-5}{4}, \frac{5}{16}, \frac{-5}{64}, ~.~.~.$
Solution:
Part (i):
1. The infinite geometric series associated with the given G.P is:
$1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$
$\begin{array}{ll}{} &{S} & {~=~} &{\frac{a}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{1}{1-\frac{1}{2}}} &{} \\
{} &{} & {~=~} &{2} &{} \\
\end{array}$
Part (ii):
1. The infinite geometric series associated with the given G.P is:
$1 - \frac{1}{2} + \frac{1}{2^2} - \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $-\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$
$\begin{array}{ll}{} &{S} & {~=~} &{\frac{a}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{1}{1- \left(-\frac{1}{2} \right)}} &{} \\
{} &{} & {~=~} &{\frac{1}{1+ \frac{1}{2}}} &{} \\
{} &{} & {~=~} &{\frac{1}{\frac{3}{2}}} &{} \\
{} &{} & {~=~} &{\frac{2}{3}} &{} \\
\end{array}$
Part (iii):
1. The infinite geometric series associated with the given G.P is:
$\frac{-5}{4} + \frac{5}{16} - \frac{5}{64} + ~.~.~.$
2. First term a is $\frac{-5}{4}$ and the common ratio r is $-\frac{1}{4}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$
$\begin{array}{ll}{} &{S} & {~=~} &{\frac{a}{1-r}} &{} \\
{} &{} & {~=~} &{\frac{\frac{-5}{4}}{1- \left(-\frac{1}{4} \right)}} &{} \\
{} &{} & {~=~} &{\frac{\frac{-5}{4}}{1 + \frac{1}{4}}} &{} \\
{} &{} & {~=~} &{\frac{\frac{-5}{4}}{\frac{5}{4}}} &{} \\
{} &{} & {~=~} &{-1} &{} \\
\end{array}$
In the next section, we will see exponential series.
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