{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

P8 - Problem 4.17 Consider the sequence{In of real numbers...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern