Infinite Series PDF

Appendix 1 INFINITE SERIES A.1.1 Introduction As discussed in the Chapter 9 on Sequences and Series, a sequence a1, a2,..., an,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a1 + a2 + a3 +... + an +... is called an infinte...

Appendix 1 INFINITE SERIES A.1.1 Introduction As discussed in the Chapter 9 on Sequences and Series, a sequence a1, a2,..., an,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a1 + a2 + a3 +... + an +... is called an infinte series associated with infinite sequence. This series can also be expressed in abbreviated form using the sigma notation, i.e., ∞ a1 + a2 + a3 +... + an +... = ∑a k =1 k In this Chapter, we shall study about some special types of series which may be required in different problem situations. A.1.2 Binomial Theorem for any Index In Chapter 8, we discussed the Binomial Theorem in which the index was a positive integer. In this Section, we state a more general form of the theorem in which the index is not necessarily a whole number. It gives us a particular type of infinite series, called Binomial Series. We illustrate few applications, by examples. We know the formula (1 + x)n = n C0 + n C1 x +... + n Cn xn Here, n is non-negative integer. Observe that if we replace index n by negative integer or a fraction, then the combinations n C r do not make any sense. We now state (without proof), the Binomial Theorem, giving an infinite series in which the index is negative or a fraction and not a whole number. Theorem The formula m ( m − 1) 2 m ( m − 1)( m − 2 ) 3 (1 + x ) m = 1 + mx + x + x +... 1.2 1.2.3 holds whenever x < 1. 2024-25  INFINITE SERIES 315 Remark 1. Note carefully the condition | x | < 1, i.e., – 1< x < 1 is necessary when m is negative integer or a fraction. For example, if we take x = – 2 and m = – 2, we obtain ( −2 )( −3) (1 − 2 )−2 = 1 + ( −2 )( −2 ) + ( −2 )2 +... 1.2 or 1= 1 + 4 + 12 +... This is not possible 2. Note that there are infinite number of terms in the expansion of (1+ x)m, when m is a negative integer or a fraction m m   b  m b ( a + b) =  a 1 +   = a 1 +  m Consider   a   a  b m ( m − 1)  2 b = a 1 + m a + 1.2   +... m  a  m −1 m ( m − 1) = a + ma b + m a m − 2b 2 +... 1.2 b This expansion is valid when < 1 or equivalently when | b | < | a |. a The general term in the expansion of (a + b)m is m ( m − 1)( m − 2 )...( m − r + 1) a m − r b r 1.2.3...r We give below certain particular cases of Binomial Theorem, when we assume x < 1 , these are left to students as exercises: 1. (1 + x) – 1 = 1 – x + x2 – x3 +... 2. (1 – x) – 1 = 1 + x + x2 + x3 +... 3. (1 + x) – 2 = 1 –2 x + 3x2 – 4x3 +... 4. (1 – x) – 2 = 1 +2x + 3x2 + 4x3 +... 1 −  x 2 Example 1 Expand  1 −  , when | x | < 2.  2 2024-25 316 MATHEMATICS Solution We have 1  1  1  3   x −  −  −x  −  −  2 = 1 +  2    +  2  2   − x  +... 2 1 −       2 1  2  1. 2  2 x 3x 2 = 1+ + +... 4 32 A.1.3 Infinite Geometric Series From Chapter 9, Section 9.5, a sequence a 1 , a2 , a 3 ,..., a n is called G.P., if ak +1 ak = r (constant) for k = 1, 2, 3,..., n–1. Particularly, if we take a1 = a, then the resulting sequence a, ar, ar2,..., arn–1 is taken as the standard form of G.P., where a is first term and r, the common ratio of G.P. Earlier, we have discussed the formula to find the sum of finite series a + ar + ar2 +... + arn – 1 which is given by Sn = ( a 1− rn ). 1− r In this section, we state the formula to find the sum of infinite geometric series a + ar + ar2 +... + arn – 1 +... and illustrate the same by examples. 2 4 Let us consider the G.P. 1, , ,... 3 9 2 Here a = 1, r =. We have 3 n 2 1−    3  = 3 1 −  2   n Sn =    ... (1) 2   3   1− 3 n  2 Let us study the behaviour of   as n becomes larger and larger.  3 2024-25  INFINITE SERIES 317 n 1 5 10 20 n 2   0.6667 0.1316872428 0.01734152992 0.00030072866 3 n 2 We observe that as n becomes larger and larger,   becomes closer and closer to 3 n 2 zero. Mathematically, we say that as n becomes sufficiently large,   becomes 3 n 2 sufficiently small. In other words, as n → ∞,   → 0. Consequently, we find that 3 the sum of infinitely many terms is given by S = 3. Thus, for infinite geometric progression a, ar, ar2,..., if numerical value of common ratio r is less than 1, then ( a 1− rn ) a − ar n Sn = = 1− r 1− r 1− r ar n In this case, r n → 0 as n → ∞ since | r | 2 and hence 2 < e < 3. Remark The exponential series involving variable x can be expressed as x x 2 x3 xn ex = 1 + + + +... + +... 1! 2! 3! n! Example 3 Find the coefficient of x2 in the expansion of e2x+3 as a series in powers of x. Solution In the exponential series x x 2 x3 e = 1+ x + + +... 1! 2! 3! replacing x by (2x + 3), we get 2024-25 320 MATHEMATICS 2 x+ 3 1+ ( 2 x + 3 ) ( 2 x + 3) + 2 +... e = 1! 2! ( 2 x + 3) n (3+ 2 x) n Here, the general term is =. This can be expanded by the n! n! Binomial Theorem as 1 n n 3 + C1 3n −1 ( 2 x ) + n C2 3n− 2 ( 2 x ) +... + ( 2 x ) . 2 n n!   2 n C 2 3n − 2 2 2 Here, the coefficient of x is. Therefore, the coefficient of x2 in the whole n! series is ∞ n C 2 3n − 2 22 ∞ n ( n − 1) 3 n− 2 ∑ n =2 n! = 2 n= 2 ∑ n! ∞ 3n –2 = 2 ∑ n = 2 ( n − 2 )! [using n! = n (n – 1) (n – 2)!]  3 32 33  = 1 + + + +... 2  1! 2! 3!  = 2e3. Thus 2e is the coefficient of x2 in the expansion of e2x+3. 3 Alternatively e2x+3 = e3. e2x  2 x (2 x ) 2 (2 x)3  e3 1 + + + +... =  1! 2! 3!  2 3 2 Thus, the coefficient of x2 in the expansion of e2x+3 is e. = 2e 3 2! Example 4 Find the value of e2, rounded off to one decimal place. Solution Using the formula of exponential series involving x, we have x x 2 x3 xn ex = 1 + + + +... + +... 1! 2! 3! n! 2024-25  INFINITE SERIES 321 Putting x = 2, we get 2 2 2 23 2 4 25 2 6 e2 = 1 + + + + + + +... 1! 2! 3! 4! 5! 6! 4 2 4 4 = 1+ 2 + 2 + + + + +... 3 3 15 45 ≥ the sum of first seven terms ≥ 7.355. On the other hand, we have  2 22 23 24  25  2 22 23  e2 < 1 + + + +  +  1 + + 2 + 3 +...   1! 2! 3! 4!  5!  6 6 6    4  1 1  4 1  2 2 = 7 +  1 + +   +...  = 7 +   = 7 + = 7.4. 15  3  3   15 1  1 −  5  3 Thus, e2 lies between 7.355 and 7.4. Therefore, the value of e2, rounded off to one decimal place, is 7.4. A.1.5 Logarithmic Series Another very important series is logarithmic series which is also in the form of infinite series. We state the following result without proof and illustrate its application with an example. Theorem If | x | < 1, then x 2 x3 log e (1 + x ) = x −+ −... 2 3 The series on the right hand side of the above is called the logarithmic series. A Note The expansion of log (1+x) is valid for x = 1. Substituting x = 1 in the expansion of log (1+x), we get e e 1 1 1 log e 2 = 1 – + – +... 2 3 4 2024-25 322 MATHEMATICS Example 5 If α, β are the roots of the equation x 2 − px + q = 0 , prove that α 2 + β2 2 α3 + β3 3 ( ) log e 1 + px + qx 2 = ( α + β ) x − 2 x + 3 x −...  α 2 x 2 α 3 x3   β 2 x 2 β3 x3  Solution Right hand side =  α x − + −... +   β x − + −...  2 3   2 3  = log e (1 + α x ) + log (1 + β x ) ( = log e 1 + ( α + β ) x + αβx 2 ) = log e (1 + px + qx ) 2 = Left hand side. Here, we have used the facts α + β = p and αβ = q. We know this from the given roots of the quadratic equation. We have also assumed that both | α x | < 1 and | β x | < 1. —v — 2024-25