SEQUENCES AND SERIES

Comparative Summary

Exercises, Convergence and Divergence of Sequences

Arithmetic Sequences Geometric Sequences A. Sequence represents an ordered list of numbers with a common difference d. Ex.: 1, 5, 9, 13, 17, 21, 25, ... , G. Sequence represents an ordered list of numbers with a common ratio r. Ex.: 1, 4, 16, 64, 256, 1024, ... , A. Series and G. Series are the values obtain by adding up all the terms of a sequence. there also called the "sum". Ex.: 1 + 5 + 9 + 13 + 17 + 21 + 25 + ... + (ASs or APs) Ex.: 1 + 4 + 16 + 64 + 256 + 1024 + ... + (GSs or GPs) Common difference is always the same value from one term to the next by adding (or subtracting) both terms. d = an+1 − an Ex: d = 9 − 5 = 4 (ASs above) Common ratio is always the same value from one term to the next by dividing (or multiplying) both terms. r = an+1 an Ex: r = 64 16 = 4 the nth number                 an = a1 + (n − 1)·d the nth number                 an = a1 · rn − 1 Sum of the nth 1st numbers                 Sn = n(a1 + an) 2 Sum of the nth 1st numbers      Sn = a1(rn − 1) r − 1 = a(1 − rn) 1 − r with a1 = a, and the special circumstance that the r is between –1 and 1, | r | < 1. the infinite sum has the following value: S∞ = a 1 − r , with | r | < 1 Sum of a Finite Geometric Series Let's find the sum of the first 7 terms of sequence S, with a1 = 1, r = 2 and n = 7. We would then plug those numbers into the formula and get: S7 = 1(1 − 27) 1 − 2 = (1 − 128) − 1 = − 127 − 1 = 127 Sum of an infinite Geometric Series An infinite number of terms a sequence S in given by: 16 + (-8) + 4 + (-2) + 1 + ..., with a1 = 16 and r = −1/2. r lies within the acceptable limits (− 1 and 1), therefore a finite sum exists when using the infinite term sum formula: S∞ = a1 1 − r = 16 1 − (−1/2) = 16 3/2 = 16 2 3 = 10.67 Convergence and Divergence of Sequences Examples Showing convergence or divergence of the the following sequence: 1, 2, 3, 4, ... Ans. This sequence starts with and goes on bigger and bigger to the infinity (∞), therefore the sequence Diverges. Does the following sequence converges or diverges? 1, 0.1, 0.01, 0.001, 0.0001, ... Ans. The sequence gets smaller and smaller and closer to 0, therefore it converges. Assuming we have the nth number of a sequence define by: an = 3(−1)n n! Ans. To show weither the sequence converges or diverges we have 2 possibilities: we can first find the 3 or 4 first terms of the sequence - a1 = 3(−1)1 1! = −3 - a2 = 3(−1)2 2! = 3 2·1 = 3 2 - a3 = 3(−1)3 3! = −3 3·2·1 = −1 2 - a4 = 1 8 ... As you can see, the terms are getting smaller and smaller and tending to 0, hence the sequence converges, (not obvious though!!!) That's why we will use limits to make it clear... lim n→∞    [ 3(−1)n n! ] = ∞ ∞    (indeterminate form!) we use L'Hôpital's rule (by derivating several times until we get the form allowing the determination of the limits) the second differentiation gives us: lim n→∞    [ 3n(−1)n−1 1! ] = ∞·0 1! = 0 Now, we can clearly see that the sequence converges, as it equals to 0 an = n(n −1) converges or diverges? lim   n(n − 1) = ∞·∞ = ∞ ⇨ the sequence diverges! n→∞

SEQUENCES AND SERIES - EXERCISES

Exercises Write down the first four terms in ascending power of x of the binomials expansion of (a)  (1 + 3x)12      (b)  (1 − 2x)9 (c)  (2 − x)10      (d)  (1 − x 3 )20      Answers (e)  (2 − 3 2x )7      (f)  ( 3 2 + 2x)9 The sum of the nth terms of a progression is given by Sn where Sn = n(3n − 4). Show that the progression is an AP.     Answer The sum of the nth terms of an AP is Sn where Sn = n² − 3n. Write down the fourth term and the nth term.     Answer Answers to i Notice: To solve these problems, we can use Pascal's Triangle, but that needs a lot of time! An alternative way is to use the Binomial Theorem, which offers a quicker method for expanding any complex binomial series. Only and only when the power n is a positive integer. see (a + b)n (a) (1 + 3x)12      = 12c0·112·(3x)0 + 12c1·111·(3x)1 + 12c2·110·(3x)2 + 12c3·19·(3x)3 + ... + (3x)12      = 1 + 36x + 594x² + 5940x3 + ... + 531441x12 (b) (1 − 2x)9      = 9c0·19·(−2x)0 + 9c1·18·(−2x)1 + 9c2·17·(−2x)2 + 9c3·16·(−2x)3 + ... + (−2x)9      = 1 − 18x + 144x² − 672x3 + ... − 512x9 (c) (2 − x)10      = 10c0·110·(−x)0 + 10c1·19·(−x)1 + 10c2·18·(−x)2 + 10c3·17·(−x)3 + ... + (−x)10      = 1024 − 5120x + 11520x² − 15360x3 + ... + x10 (d) (1 − x 3 )20      = 20c0·120·(−x/3)0 + 20c1·119·(−x/3)1 + 20c2·118·(−x/3)2 + 20c3·117·(−x/3)3 + ... +      = 1 − 20x/3 + 190x²/9 − 1140x3/27 + ... + x20/320 (e) (2 − 3 2x )7      = 7c0·27·(−3/2x)0 + 7c1·26·(−3/2x)1 + 7c2·25·(−3/2x)2 + 7c3·24·(−3/2x)3 + ... + ...      = 128 − 672/x + 1512/x² − 1890/x3 + ... + ... (f) ( 3 2 + 2x)9      = 9c0·(3/2)9·(2x)0 + 9c1·(3/2)8·(2x)1 + 9c2·(3/2)7·(2x)2 + 9c3·(3/2)6·(2x)3 + ... +      = (3/2)9 + (310/27)x + (39/8)x² + (7·(37)/2)x3 + ... + ... Answer to ii the first nth terms are given by an = Sn − Sn−1 ⇒ an = n(3n − 4) − (n − 1) [3(n − 1) − 4] = 3n² − 4n − (n − 1) (3n − 7) = 3n² − 4n − 3n² + 10n − 7) ⇒ an = 6n − 7, therefore, we can find out:       a1 = 6·1 − 7 = − 1 ⤸+6       a2 = 6·2 − 7 = 5 ⤸+6       a3 = 6·3 − 7 = 11 ⤸+6       a4 = 6·4 − 7 = 17 ⤸+6       ...      ...      ...       ...      ...      ...       an = 6·n − 7       and the sequence is: − 1, 5, 11, 17, 23,..., As you can see, the common difference d = 6. By replacing a1 and an in the AP's general formula, we should obtain Sn = n(3n − 4), which is the proof that Sn is an AP progression. By replacing a1 and an in the AP's general formula, we should obtain Sn = n(3n − 4), which is the proof that Sn is an AP progression. Sn = n(a1 + an) 2 = n(−1 + 6n − 7) 2 = 2n(3n − 4) 2 ⇒ Sn = n(3n − 4), the progression is an AP. Answer to ii * The nth term is given by: an = Sn − Sn−1 ⇒ an = n² − 3n − [(n−1)² − 3(n−1)] = n² − 3n − (n² − 5n + 4) = n² − 3n − n² + 5n − 4 ⇒ an = 2n − 4 * The 4th term a4 = 4 Find the first three terms in the expansion of (1 + b)5. Use the substitution x = 1 + b, to find the coefficient of b² in the expansion of (1 + b − 2b²)5. Answer ⤵ * To expand (1 + b)5, we use the Binomial Theorem: ∴ (1 + b)5 = 5C0(1)5(6)0 + 5C1(1)4(6)1 + 5C2(1)3(6)2 + ... + 5C5(1)0(6)5 = 1 + 5 + 10 ⇒ (1 + b)5 = 1 + 5 + 10, which are the 1st three terms of the expansion. * Now x = 1 + b is plugged in the expression (1 + b − 2b²)5 by substitution ⇒ (x − 2b²)5. We can still use the Binomial Theorem, but let's go for the Pascal's Triangle. 0 1 1      1       1 2 1        2 1 3 1    3      3        1 4     1    4 6    4     1 5       1    5   10       10           5   1 As you can see, we're concerned by the 5th line because our power is 5 ∴ we can write: 1(x)5(−2b²)0 + 5(x)4(−2b²)1 + 10(x)3(−2b²)2 + 10(x)2(−2b²)3 + 5(x)1(−2b²)4 + 1(x)0(−2b²)5 and after working out the parenthesis we obtain: x5 − 10x4b² + 40x3b4 − 80x²b6 + 80xb8 − 32xb10. Then, the coefficient of b² is − 10 Close ⤴ The coefficient of b² and b³ in the expansion of (a + b)6 are equal. Find the value of a.