Test 7 - Sol

Pure Mathematics – Sequences

p.1

King’s College 2007 – 2008 F.6 Pure Mathematics Revision Test 7 Suggested Solution

1.

When n = 1, x1 = 2 > 1 and 1 −

(a) (i)

1 1 1 1 = 1− = = = S1 . x2 − 1 3 − 1 2 x1

∴ The statements are true for n = 1.

Assume that x k > k and S k = 1 −

1 x k +1 − 1

Then,

Then, x k +1 = x k − x k + 1

x k +1 = x k − x k + 1

2

= x k (x k − 1) + 1 > k ( x k − 1) + 1

∴

x k +1

for some k ∈ N .

2

= (x k − 1) + x k 2

(Q x k

> k ≥ 1)

≥ k +1 > k +1

> xk

Since xn is an integer for all n ∴ x k +1 ≥ x k + 1 > k + 1

(ii) Also, S k +1 = S k + = 1− = 1− = 1−

1 x k +1 1

x k +1 − 1

+

1 x k +1

1 x k +1 − x k +1 2

1 xk +2 − 1

∴By mathematical induction, the statements are true for any positive integer n.

(b) By (a)(i), xn > n > 0 => {Sn} is strictly increasing. 1 < 1 => {Sn} is bounded above by 1. Also, by (a)(ii), S n = 1 − x n +1 − 1 Thus, lim S n exists. n→∞

By (a)(i), xn > n => xn Æ ∞ as n Æ ∞ ∴

⎛ 1 ⎞ ⎟⎟ = 1 → 0 as n → ∞ ⇒ lim S n = lim⎜⎜1 − n →∞ n →∞ x n +1 − 1 ⎝ x n +1 − 1 ⎠ 1

lim S n exists (and equals 1) n→∞

Pure Mathematics – Sequences

2.

p.2

(a) Let f(x) = x – ln(x + 1) for all x > -1. Then, we have f ' (x ) = 1 −

1 x +1

x x +1 ⎧< 0 if − 1 < x < 0 ⎪ ⎨= 0 if x = 0 ⎪> 0 if x > 0 ⎩

=

Therefore, f attains its absolute minimum at 0. Thus, we have f ( x ) ≥ f (0 ) for all x > -1 x − ln ( x + 1) ≥ 0 for all x > -1

x ≥ ln ( x + 1) for all x > -1

(b) Let n be a positive integer. Putting x =

1 in (a), n

1 ⎛ n +1⎞ ⎛1 ⎞ ≥ ln⎜ + 1⎟ = ln⎜ ⎟. n ⎝ n ⎠ ⎝n ⎠ For all positive integers k, k n +1 1 ≥ ln ∑ ∑ n n =1 n n =1 k n +1 = ln Π n =1 n k +1 = ln n ∞ k k +1 1 1 = = lim =∞ lim ∑ ∑ k →∞ k →∞ n n =1 n n =1 n k

∞

1

∑n

Thus, the series

is divergent.

n =1

3.

(a) (i)

an+1 – bn+1 2 a n bn a n + bn − a n + bn a n + bn 2

= =

2

(a n − bn )2 a n + bn

≥ 0 for all n = 1,2,3,K. Thus, an ≥ bn for all n = 1, 2, 3, …. − bn (a n − bn ) a n + bn ≤ 0 for all n = 1, 2, 3, …. − an = a n + bn a n + bn 2

(ii) a n +1 + a n =

2

Pure Mathematics – Sequences

p.3

Therefore, bn ≤ bn +1 for all n = 1, 2, 3, …. Thus, {bn} is monotonic increasing. b (a − bn ) 2a n bn bn +1 + bn = − bn = n n ≥ 0 for all n = 1, 2, 3, …. a n + bn a n + bn Therefore, a n ≥ a n +1 for all n = 1, 2, 3, …. Thus, {an} is monotonic decreasing. (iii) By (a)(i) and (a)(ii), a1 ≥ a n ≥ bn ≥ b1 for all n = 1, 2, 3, … {an} is monotonic decreasing and bounded below by b1. {bn} is monotonic increasing and bounded above by a1. l1 + l 2 l1 + l 2 2

(iv) Let lim a n = l1 and lim bn = l2. Then, we have l1 = n →∞

n→∞

2

(or l1 =

(l1 – l2) l2 = 0 But since l 2 = lim bn ≥ b1 > 0 , we have l1 = l2. n →∞

i.e. lim a n = lim bn n→∞

n→∞

(v) Note that a n + bn =

(a n−1 + bn−1 )2 a n −1 + bn −1

= a n −1 + bn −1 = K = a1 + b1 .

Therefore, we have lim(a n + bn ) = a1 + b1 . n →∞

a +b 1 1 1 lim a n + lim bn = lim(a n + bn ) = 1 1 → ∞ → ∞ → ∞ n n n 2 2 2 2 If a1 ≤ b1 < 0 , then − a1 ≥ −b1 > 0 .

By (a)(iv), lim a n = n→∞

(b)

An + Bn An + Bn 2

Define An +1 =

2

and Bn +1 =

2 An Bn for all n = 1, 2, 3, …, An + Bn

where A1 = –a1 and B1= –b1. Note that A1B1 > 0 and A1 ≥ B1 > 0 . Then, by (a), lim An and lim Bn both exist. n→∞

n→∞

Further note that an = –An and bn= –Bn for all n = 1, 2, 3, …. Therefore, we have lim a n = − lim An and lim bn = − lim Bn . n→∞

n →∞

n→∞

n →∞

Thus, the limits of the sequemces {an} and {bn} both exist. You may repeat Part(a)(i), (ii) and (iii) as an alternative method.

2l1l 2 ). l1 + l 2