In the previous section, we completed a discussion on probability. In this appendix A, we will see Infinite series.
• In chapter 9, we discussed sequences and series.
1. Consider the sequence given below:
$a_1, a_2, a_3,~.~.~.~,~a_n,~.~.~.$
• This sequence has infinite number of terms. So it is called an infinite sequence.
2. Let us write the corresponding series:
$a_1~+~a_2~+~a_3~+~.~.~.~,~+~a_n~+~.~.~.$
• This series is called the infinite series associated with the infinite sequence.
3. We can abbreviate an infinite series as shown below:
$a_1~+~a_2~+~a_3~+~.~.~.~,~+~a_n~+~.~.~.~=~\sum\limits_{i=1}^{i=\infty} a_i$
• In chapter 8, we discussed binomial theorem. Based on that theorem, we derived the formula:
$(1+x)^n~=~{}^n \rm{C}_0~+~{}^n \rm{C}_1 x~+~{}^n \rm{C}_2 x^2~+~{}^n \rm{C}_3 x^3~+~.~.~.~+~{}^n \rm{C}_n x^n$
• "n" in the above formula must be an integer. Also, it should not be -ve. This is because, in such cases, we will not be able to calculate ${}^n \rm{C}_r$.
• We will now see a formula which can be used when n is not an integer and/or n is -ve. It can be explained in 4 steps:
1. The formula is:
$(1+x)^m~=~1~+~mx~+~{\frac{m(m-1)}{1 \times 2}}x^2~+~{\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}x^3~+~{\frac{m(m-1)(m-2)(m-3)}{1 \times 2 \times 3 \times 4}}x^4~+~.~.~.$
• We will see the derivation of this formula in higher classes.
2. This formula is applicable whenever |x| < 1
• This condition can be explained in 5 steps:
(i) Consider the value of x. Based on that value, we can mark it’s position on the number line.
(ii) The distance of this mark from zero must be less than 1.
(iii) So, if x is -ve,
♦ The mark must not be on -1.
♦ The mark must not be anywhere to the left of -1.
(iv) Similarly, if x is +ve,
♦ The mark must not be on 1.
♦ The mark must not be anywhere to the right of 1.
(v) We can combine the above four points as:
-1 < x < 1
• The importance of this condition will become clear when we see an example. Let us put “-3” in the place of x and "-2" in the place of m. We get:
$\begin{array}{ll}{} &{(1-3)^{-2}} & {~=~} & {1 + (-2)(-3) + \frac{(-2)(-3-1)}{1 \times 2} (-3)^2 + \frac{(-2)(-3-1)(-3-2)}{1 \times 2 \times 3} (-3)^3 + . . .} &{} \\
{\Rightarrow} &{(-2)^{-2}} & {~=~} & {1 + 6 + \frac{(-2)(-4)}{2} (9) + \frac{(-2)(-4)(-5)}{6} (-27) + . . .} &{} \\
{\Rightarrow} &{(-1)^{-2} \times (2)^{-2}} & {~=~} & {1 + 6 + \frac{8}{2} (9) + \frac{-40}{6} (-27) + . . .} &{} \\
{\Rightarrow} &{1 \times \frac{1}{4}} & {~=~} & {1 + 6 + (4) (9) + \frac{-20}{3} (-27) + . . .} &{} \\
{\Rightarrow} &{\frac{1}{4}} & {~=~} & {1 + 6 + 36 + 180 + . . .} &{} \\
\end{array}$
• This is not possible. So before applying this formula, we must make sure that -1 < x < 1.
3. We know how to expand (a+b)m when m is a +ve integer. But what if m is not an integer and/or -ve?
In such cases, we can use the above formula. This is shown below:
$\begin{array}{ll}{} &{(a+b)^m} & {~=~} & {\left[a \left(1 + \frac{b}{a} \right) \right]^m} &{} \\
{} &{} & {~=~} & {a^m \left(1 + \frac{b}{a} \right)^m} &{} \\
{} &{} & {~=~} & {a^m \left[1 + m{\frac{b}{a}} + {\frac{m(m-1)}{1 \times 2}}\left(\frac{b}{a} \right)^2 + {\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}\left(\frac{b}{a} \right)^3~+~.~.~. \right]} &{} \\
{} &{} & {~=~} & {a^m + m a^{m-1} b + {\frac{m(m-1)}{1 \times 2}}a^{m-2} b^2 + {\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}a^{m-3} b^3 ~+~.~.~.} &{} \\
\end{array}$
• We were able to use the formula because we wrote (a+b)m as $\left[a \left(1 + \frac{b}{a} \right) \right]^m$.
• Recall that, to use the formula, |x| must be less than 1.
• So in our present case, $\left| \frac{b}{a} \right| $ must be less than 1.
• This can be explained in 5 steps:
(i) Values of a and b:
♦ a can be +ve or -ve. It can be any real number.
♦ b can be +ve or -ve. It can be any real number.
(ii) Calculating $\frac{b}{a}$:
We calculate $\frac{b}{a}$ using the proper signs.
(iii) Marking the position on the number line:
We mark $\frac{b}{a}$ based on the result in (ii).
(iv) The mark can be either on the left side or right side of zero. But it's distance from zero must be less than 1.
Let us see some examples:
Example1:
• Put a = 2 and b = -3
• Then $\frac{b}{a} = \frac{-3}{2}$ = -1.5
• When we mark $\frac{b}{a}$ on the number line, it will be at a distance of 1.5 units from zero. So we will not be able to use the formula.
Example2:
• Put a = 4 and b = -3
• Then $\frac{b}{a} = \frac{-3}{4}$ = -0.75
• When
we mark $\frac{b}{a}$ on the number line, it will be at a distance of
0.75 units from zero. So we can use the formula
(v) Based on the above four steps, we can write:
$\frac{b}{a}$ must be a proper fraction. It can be +ve or -ve.
4. Based on the expansion in (3), we can write the general term in the expansion. It is given below:
$$\frac{m(m-1)(m-2)(m-3)~.~.~.~(m-r+1) a^{m-r} b^r}{1 \times 2 \times 3 \times ~.~.~.\times r}$$
Based on the formula that we wrote in (1), we can derive four useful results:
Result I:
$\begin{array}{ll}{} &{(1+x)^{-1}} & {~=~} &{1 + (-1)x + \frac{(-1)(-1-1)}{1 \times 2} x^2 + \frac{(-1)(-1-1)(-1-2)}{1 \times 2 \times 3} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-1)x + \frac{(-1)(-2)}{1 \times 2} x^2 + \frac{(-1)(-2)(-3)}{1 \times 2 \times 3} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-1)x + \frac{(-1)(-1)}{1 \times 1} x^2 + \frac{(-1)(-1)(-1)}{1 \times 1 \times 1} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 - x + x^2 – x^3~+~.~.~.} &{} \\
\end{array}$
Result II:
$\begin{array}{ll}{} &{(1-x)^{-1}} & {~=~} &{[1+(-x)]^{-1}} &{} \\
{} &{} & {~=~} &{1 + (-1)(-x) + \frac{(-1)(-1-1)}{1 \times 2} (-x)^2 + \frac{(-1)(-1-1)(-1-2)}{1 \times 2 \times 3} (-x)^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-1)(-x) + \frac{(-1)(-2)}{1 \times 2} (x^2) + \frac{(-1)(-2)(-3)}{1 \times 2 \times 3} (-1)(x^3)~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-1)(-x) + \frac{(-1)(-1)}{1 \times 1} (x^2) + \frac{(-1)(-1)(-1)}{1 \times 1 \times 1} (-1)(x^3)~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + x + x^2 + x^3~+~.~.~.} &{} \\
\end{array}$
Result III:
$\begin{array}{ll}{} &{(1+x)^{-2}} & {~=~} &{1 + (-2)x + \frac{(-2)(-2-1)}{1 \times 2} x^2 + \frac{(-2)(-2-1)(-2-2)}{1 \times 2 \times 3} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-2)x + \frac{(-2)(-3)}{1 \times 2} x^2 + \frac{(-2)(-3)(-4)}{1 \times 2 \times 3} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-2)x + \frac{(-1)(-3)}{1 \times 1} x^2 + \frac{(-1)(-1)(-4)}{1 \times 1 \times 1} x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 - 2x + 3 x^2 –4 x^3~+~.~.~.} &{} \\
\end{array}$
Result IV:
$\begin{array}{ll}{} &{(1-x)^{-2}} & {~=~} &{[1+(-x)]^{-2}} &{} \\
{} &{} & {~=~} &{1 + (-2)(-x) + \frac{(-2)(-2-1)}{1 \times 2} (-x)^2 + \frac{(-2)(-2-1)(-2-2)}{1 \times 2 \times 3} (-x)^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-2)(-x) + \frac{(-2)(-3)}{1 \times 2} (-x)^2 + \frac{(-2)(-3)(-4)}{1 \times 2 \times 3} (-x)^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + (-2)(-x) + \frac{(-1)(-3)}{1 \times 1} x^2 + \frac{(-1)(-1)(-4)}{1 \times 1 \times 1} (-1) x^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + 2x + 3x^2 + 4x^3~+~.~.~.} &{} \\
\end{array}$
Now we will see a solved example:
Solved example 1
Expand $\left(1 - \frac{x}{2} \right)^{- \frac{1}{2}}$, when |x| < 2.
Solution:
1. We can rewrite the given expression as:
$\left[1 + \left(- \frac{x}{2} \right) \right]^{- \frac{1}{2}}$
• Consider the term $\left(- \frac{x}{2} \right)$.
We know that, this term must be less than 1 and at the same time, greater than -1. This condition will be satisfied only if |x| is less than 2. The reader must do the necessary analysis and become convinced about this fact.
2. Now we can write the expansion:
$\begin{array}{ll}{} &{\left[1 + \left(- \frac{x}{2} \right) \right]^{- \frac{1}{2}}} & {~=~} &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{\left(-\frac{1}{2} \right)\left(-\frac{1}{2} – 1 \right)}{1 \times 2} \left(-\frac{x}{2} \right)^2 + \frac{\left(-\frac{1}{2} \right)\left(-\frac{1}{2} – 1 \right)\left(-\frac{1}{2} – 2 \right)}{1 \times 2 \times 3} \left(-\frac{x}{2} \right)^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{\left(-\frac{1}{2} \right)\left(-\frac{3}{2} \right)}{1 \times 2} \left(-\frac{x}{2} \right)^2 + \frac{\left(-\frac{1}{2} \right)\left(-\frac{3}{2} \right)\left(-\frac{5}{2} \right)}{1 \times 2 \times 3} \left(-\frac{x}{2} \right)^3~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{3}{8} \left(-\frac{x}{2} \right)^2 + \frac{-5}{16} \left(-\frac{x}{2} \right)^3 ~+~.~.~.} &{} \\
{} &{} & {~=~} &{1 + \frac{x}{4} +\frac{3 x^2}{32} + \frac{5 x^3}{128} ~+~.~.~.} &{} \\
\end{array}$
In the next section, we will see infinite geometric series.
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