{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3100_Homework9_sols

3100_Homework9_sols - SOLUTIONS TO PROBLEM SET 9 PETE L...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTIONS TO PROBLEM SET 9 PETE L. CLARK All problems except the last are taken from § 2 . 3 of our text. 1) Let { s n } be a sequence and { c n } and { d n } be subsequences given by c n = s 2 n and d n = s 2 n +1 . Assume that lim n →∞ c n = lim n →∞ d n = L . Show that lim n →∞ s n = L . Solution: Let ϵ > 0. Since the sequences { c n } and { d n } each converge to L , there exists N 1 N such that for all n N 1 , | c n - L | < ϵ and N 2 N such that for all n N 2 , | d n - L | < ϵ . Let N = 2 max( N 1 , N 2 ) + 1. Any n N may be written as 2 k or 2 k + 1 with k max( N 1 , N 2 ). If n = 2 k , then | s n - L | = | s 2 k - L | = | c k - L | < ϵ, whereas if n = 2 k + 1, then | s n - L | = | s 2 k +1 - L | = | d k - L | < ϵ. Thus s n L . Remark: More generally if we partition the natural numbers N into any finite collection of infinite sets S 1 . . . S K , then a sequence s n L iff for all 1 i k , the subsequence obtained by restricting the terms to lie in S i converges to L . 2) What does the ratio test tell you about the following series? a) n 2 n n ! . Solution: The ratio test limit is ρ = lim n →∞ 2 n +1 ( n + 1)! n ! 2 n = lim n →∞ 2 n + 1 = 0 < 1 , so the series converges. b) n n ! 2 n . Solution: Since the terms of this series are the reciprocals oft he terms of the series in part a), the ratio test limit ρ will be the reciprocal of the ρ for part a), i.e., ρ = 1 0 + = . So the series diverges. c) n n 3 n . c Pete L. Clark, 2011. 1
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
2 PETE L. CLARK Solution: The ratio test limit is ρ = lim n →∞ n + 1 3 n +1 3 n n = lim n →∞ 1 3 n + 1 n = 1 3 < 1 , so the series converges. d) n 1 n 2 . Solution: The ratio test limit is ρ = lim n →∞ n 2 ( n + 1) 2 = 1 , so the ratio test fails , i.e., it does not tell us whether the series converges. In fact this is a p -series with p = 2 > 1, so by other means – the Condensation Test or the Integral Test – we know the series converges. So it is not the convergence of the series that is failing, it’s the Ratio Test! 3) Which of the following series are absolutely convergent, nonabsolutely conver- gent or divergent? a) n ( - 1) n n 2 . Solution: The associated series of absolute values is the convergent p -series n 1 n 2 , so the given series is absolutely convergent.
Image of page 2
Image of page 3
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