Unformatted text preview: Problem 4.17. Consider the sequence {In} of real numbers deﬁned by x1 = l
and x,.+1 = l + 1—17 for n = 1.2.... . Show that {In} is a convergenr sequence and that lim 1,. = ﬂ. Solution. _l"l1'1ea.'~=3.~inchilctiveargunumtxshowsthaws,l > Oforeech n. This implies that, in fact, we have 1 5 x... 5 2 for each n. Now. note that l 1 1+1" _ 1+In_1 ﬂ lxu — xn—li
" (1+x.)u+x.._o
< ixn—xn—li 3%
"' (1 + 1)(1 + 1) for each n = 2, 3, . . . . By Problem 4.15, the sequence {.15.} converges. Let held 314 = I ixn 4 xu—li x" w) x. Sineexn a 1 foreachn.weseeﬂ1atx 2 E. Then i
)=1+1+I. x = linen“ =.1::e(1 + i..." That is, x is the positive solution of the equationx = 1+ i . or x2+x = 1+x+ 1.
This implies J:2 = 2. and so I = 1/5. Problem 4.18. Deﬁne the sequence [xn] by I. = l and x,,+1=%(x,,+—2—). r1:1.2,.... J:n
Show that [1,.) converges and that lire 1,. = «5. Solution. Clearly, .15.. > 0 holds for each 1:. (Use induction to prove this!) Also, and so!) < x,,.H < x" holds for each n 2 2. By Theorem 4.3,x m limx" exism.
Since 13' 2 2 holds for each n z 2, we see that x > 0. From the recursive
formula, it follows that 21 = x + f, or x2 = 2. (Note also that the limit is
independent of the iniu'al choice I] > 0.) Problem 4.19. Deﬁne the sequence x" = ELI % ﬁg: n = l, 2, . , . . Show that
{xn} does no! converge in IR. (See also Problem 5.10.) Solution. The inequality “"1... .J_ L Ina—1'" — n+1 +n+1+ +n+n
1 l _ lei
35+E+”'+ﬁ—"'h“2 shows that {In} is not a Cauchy sequence. and hence, is not convergent in R. ...
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 Spring '10
 GuoliangWu
 Math

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