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# P6 - Problem 4.12 For a sequence ofteal numbers{xn...

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Unformatted text preview: Problem 4.12. For a sequence ofteal numbers {xn} establish thefollawing: a. If x“: "x" -9 x in R. then 1..."): —r x. b. [ftxﬂ] is bounded and 2x" 5 x“; +x,,_l holdsfor alln = 2.3. . . . . then xn+1 ... x" T 0. Solution. (a) Assume that x”. —x,. ~+ x in R. Noticethat 22:31..“ "1.) = x...“ — x1 for each :1. By Problem 4.11, we have ﬁgﬂﬁn -Zt) = “1.1+: —XI) ‘4’ I- Since xmt —> 0. it follows that x.“ /n ~> x. New note that x,I Jt,l n -—1 n 21-1 in —>x-l=x. (b)'I'he condition 2):, 5 1".“ + x34 can be rewritten as 1,. — x,,_l 5 x..+1 — x" for each n = 2, 3. . . .. which implies that unbounded sequence [1,,“ —x,,] isan increasing sequence, and hence convergent. Let x.“ — x" t x in R. By part (a). we have In]?! -—> x. But. since [x,.] is abounded sequence, xu/n —> 0. Therefore, x =0,andsox..+l -x,. 1‘0. Problem 4.13. Consider the sequence [x,.) deﬁned by 0 < x, < 1 and x,.+1 = 1—1/l—xnforn =1.2..... Showthat x,‘ l, 0.A£ro,rhowthat it? —> % Solution. We claim that 0<x,.+. <Jtll <1 (‘0 holds for each n = 1, 2, . To verify this claim, we use induction. Since 0 <11 < l.wehnve0< l -:c. < Landsat) < l —x; < Jlmxl <1. Hence. 0 < 1—- ‘/l -x. =.r2 <1. <1.‘I'h£ttis.(*)istrueforn = 1. For the. inductive argument. assume that (*) is true for some rt. This implies 0< l—x" <1—x,,+1<1.ands00 < ./1—x,, < ‘/I—-x,,+1<l.fromwhich itfollowsthat 0<xn+2=1—1/1—x,.+1 <1—vl—x}.=:r,.+1 <1, which shows that (s) is true for n + 1. This completes the induction and guarantees that (n) is true for each n. Now, since [xn] is decreasing and bounded from below. it converges, say to x e R. Clearly,0 S x < 1. Moreover. we have x=ﬂl_i’nat°x,.+l = lim(1—\/1—x,,)=1—Jl—x. n-roo In other words, x is the non-negative solution of the equation x = 1 - J1 — . Solving the equation yields x = 0 or .t = 1. Hence, I = 0, and so x" L 0. For the last part, notice that xn+l_i— ‘xn 1 _—gm_., In In 1+IJI—xn and the solution is complete. ...
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