Tuesday, October 3, 2023

A.1 Infinite Geometric Series

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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