{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat25-hw8solns - Math 25 Advanced Calculus UC Davis Spring...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 – Solutions to Homework Assignment #8 1. (a) lim sup n →∞ a n = 1; lim inf n →∞ a n = - 1; the set of subsequential limits is {- 1 , 1 } . (b) lim sup n →∞ a n = 1; lim inf n →∞ a n = - 1; the set of subsequential limits is {- 1 , - 2 - 1 2 , 0 , 2 - 1 2 , 1 } . (c) lim sup n →∞ a n = 1; lim inf n →∞ a n = - 1; the set of subsequential limits is [0 , 1]. (d) lim sup n →∞ a n = lim n →∞ a n = lim inf n →∞ a n = e ; the set of subsequential limits is { e } . (e) lim sup n →∞ a n = lim n →∞ a n = lim inf n →∞ a n = 1; the set of subsequential limits is { 1 } . (f) lim sup n →∞ a n = lim n →∞ a n = lim inf n →∞ a n = ; the set of subsequential limits is . 2. (a) For any m N , since { a m +1 , a m +2 , . . . } ⊆ { a m , a m +1 , a m +2 , . . . } , b m +1 = sup { a m +1 , a m +2 , . . . } ≤ sup { a m , a m +1 , a m +2 , . . . } = b m . Thus, ( b m ) m =1 is nonincreasing. (b) Since (from above) ( b m ) m =1 is nonincreasing, either there exists some k such that b n = α for some α and all n k , or not. If this is the case, then lim n →∞ b n = L = α , so that b m
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the 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