Analysis17and18 - Real Analysis II Section 17 and 18 David...

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

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Real Analysis II: Section 17 and 18 David Joseph Stith 17.6 Exercise. For each of the following prove or give a counterexample. (a) The statement, “If ( s n ) and ( t n ) are divergent sequences, then ( s n + t n ) diverges,” is false. Consider the sequences ( s n ) and ( t n ) defined by s n = − n and t n = n . Then ( s n ) and ( t n ) are divergent since ( s n ) and ( t n ) are unbounded but s n + t n = 0 for all n ∈ N so that ( s n + t n ) converges to zero. (b) The statement, “If ( s n ) and ( t n ) are divergent sequences, then ( s n t n ) converges,” is false. Consider the sequences ( s n ) and ( t n ) defined by s n = braceleftbigg − 1 , if n is odd; 1 , if n is even. and t n = s n for all n ∈ N . Then s n and t n both diverge, as has been previously shown, but s n t n = 1 for all n ∈ N so that ( s n t n ) converges to 1. (c) The statement, “If ( s n ) and ( s n + t n ) are convergent sequences then ( t n ) con- verges,” is true. Suppose ( s n ) and ( t n ) are sequences such that ( s n ) converges and ( s n + t n ) converges. Let s = lim( s n ) and u = lim( s n + t n ). Now t n = ( s n + t n ) − ( s n ), so that by Theorem 17.1(a) and 17.1(b), lim(( s n + t n ) − ( s n )) = lim( s n + t n ) − lim( s n ) since both are both convergent = u − s Therefore ( t n ) converges, and in particular, lim( t n ) = lim( s n + t n ) − lim( s n ). Q.E.D. (d) The statement, “If ( s n ) and ( s n t n ) are convergent sequences, then ( t n ) con- verges,” is false. Consider sequences ( s n ) and ( t n ) defined by s n = 0 and t n = n . Now ( s n ) converges, and s n t n = 0 for all n ∈ N so that ( s n t n ) converges, nevertheless ( t n ) is divergent since ( t n ) is unbounded. 17.7 Exercise. Give an example of an unbounded sequence that does not diverge to + ∞ or to −∞ . Consider the sequence ( s n ) defined by s n = ( − 1) n n . That ( s n ) is unbounded is evidenced by the fact that | s n | = n for all n ∈ N . Hence for any M ∈ R , | s n | > M for all n > M . But ( s n ) does not diverge to + ∞ since for any M ∈ R , s n < M for all odd n > M , nor does ( s n ) diverge to −∞ since for any M ∈ R , s n > M for all even n > M . 1 Real Analysis II: Section 17 and 18 David Joseph Stith 17.8 Exercise. (a) Give an example of a convergent sequence ( s n ) of positive numbers such that lim( s n +1 /s n ) = 1 . Consider the sequence ( s n ) defined by s n = 1 n . Now s n is positive for all n ∈ N and ( s n ) is convergent (see Example 16.3). We will show that lim( s n +1 /s n ) = 1 by showing that for any ǫ > 0, there exists N ∈ R such that for n ∈ N , n > N = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle s n +1 s n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ Suppose ǫ > 0. Then let N = 1 ǫ . Now, n > N = ⇒ n > 1 ǫ = ⇒ n + 1 > 1 ǫ = ⇒ 1 n + 1 < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle 1 n + 1 vextendsingle vextendsingle vextendsingle...
View Full Document

  • Fall '08
  • Akhmedov,A
  • Mathematical analysis, Tn, Dominated convergence theorem, David Joseph Stith

{[ 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