Question

5. Let {xn} and {yn} be sequences of real numbers such that x1 = 2 and y1 = 8 and for n = 1,2,3,···

x2nyn + xnyn2 x2n + yn2 xn+1 = x2 + y2 and yn+1 = x + y .

nn nn
(a) Prove that xn+1 − yn+1 = −(x3n − yn3 )(xn − yn) for all positive integers n.

(xn +yn)(x2n +yn2) (b) Show that 0 < xn ≤ yn for all positive integers n.

Hence, prove that {xn} is a monotonic increasing sequence and {yn} is a monotonic decreasing sequence.

(c) Prove that {xn} and {yn} converge and lim xn = lim yn. n→∞ n→∞

(d) Prove that xnyn is a constant and hence find lim xn. n→∞

6 5. Let {n} and {yn} be sequences of real numbers such that x1 = 2 and yi = 8 and x yn + XnYn and x2 + y2 Yn+1 * In + Yn - - ynxn - Yn) (a) Prove that Xn+1 - Yn+1 = for all positive integers n. (In + yn)(x + yn) (b) Show that 0 < xn syn for all positive integers n. Hence, prove that {n} is a monotonic increasing sequence and {yn} is a monotonic decreasing sequence. (C) Prove that {n} and {yn} converge and lim xn = lim Yn. (d) Prove that XnYn is a constant and hence find lim xn. non-00 n->00

Answer 1

Given {n} and [In} sequences are as ay = 2,4,= 8 de Ant- won Yu+xmin? 1 2 + - Intl = tn tin , not. kutin knep Into = xu Ynt any antyna - femint xen xom) (en + ) - (24 ayany? n + 21, 2, 4,ได้ 16, 2, -2, -, 1 (IntYn) (auty --x + kn?ynt knynya Can Ym) (xutym?) = -xx (Xn-) + you (**-*) (An tYm) (xhtm) (An-Yn) (-Xn?+Yn?) (Intyn) (xntyn) - - (xn-Ym?) (xn - Yn) (antyn) (auty , for all ne So, from (a), it follows! Katis Yuri if and only if xn < and 4 <Y. So, by mathematical induction method, we have. Dr <In for all NEIN, b)

Also, Xati - Kuntant Xn tyn show that Intrants >o if and only if xn, In> and for nal , 4 ,Y, >o ras x=2, Y,=8] olan Cyn. Now xhel-xnxñant ani n a - * , **** - - 21% n +Y Xryn-x3 anten so [as yna xn] So, xnx, xn and {an} is monotonic increasing. Also, Intilna - Antyn Yn -art yn -xniny Xn+Yn xnxrin xn (An-Y). Xntyn Intln c) So, Intl < Yn and {n} is monotonic decreasing Thes, {r} is monotonic increasing and bounded above by Y. So, it is convergent. And fin} is monotonic decreasing and bounded below by M. So, it is convergent.

9 > DO het, lim anal and Then, from xn Yn + kita lim tonal, Xnti- we have = xuntln ixteen > 1²+11~= [ltler > 1²e e te lo 7 ľ(1-1)=0 Nows e to, as all xn 2x4=2 for NEIN. Ther, e-l'zo l=1 3 limon Xin = hiyo ni hoc d) Now, Knol Y no 1= x ñ Yn tanim antyna xn't yn Xnt Yn = Krin (antyn) Intyn for all neln. So, ana Imera Knin = Xn-Yn anna2% 23 247, = 2.8 = 16 Thus, {retn} is a constant sequence.

Now, as (an} & {yn} are Convergent, exist positive real numbers K,K' such 1 kn/SK fe Mal SK , for all nflN. there that then /Xnxn-LL! = 1 an (Yu-L') +$(2n-1) Elanllyn - llt lillan-ll' Sklyn-lt lillkall het to be given. Since, lim xual de l'im hal there exist positive integers m, ma such that ut for all msm, and 1 Yn - "I < I for all n= m2. So, for m= max {m, ma}, 1 anya-lll < 1 € + 1lt 12 € Lt 16 for all nəm. Thus, lima kan Yen = el=& [ Ass.l=1] 16= ľ x=4 [ o as all this ton >o]