ans0909

ans0909 - Homework answers week of Sept 9 1 Proof#1 Since x...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Homework answers, week of Sept. 9 1. Proof #1. Since { x n } diverges to infinity, then given B > 0, there exists N > 0 such that x n > B for all n > N . The hypotheses imply that y n ≥ x n > B for all n > N . Hence { y n } → ∞ . Proof #2. Suppose that { y n } were bounded. Then { y n } has a least upper bound, say B . Therefore, x n ≤ y n ≤ B for all n , which implies that { x n } is bounded, a contradiction. Hence { y n } does not have an upper bound. 2. The problem doesn’t ask for a rigorous proof, just a determination of whether the given sequence is Cauchy. I have indicated a sample justification for each answer. (a) Problem 2, p. 81. The sequence x n = ∑ n i = 1 ( 1 / i ! ) is Cauchy. One pos- sible rationale: | x n + k- x n | < k ∑ j = 1 1 ( n + 1 ) j = 1 n 1- 1 ( n + 1 ) k < 1 n . Given ε , we may choose n so that | x n + k- x n | < ε for all k ∈ N . Hence the sequence is Cauchy....
View Full Document

{[ snackBarMessage ]}

Page1 / 2

ans0909 - Homework answers week of Sept 9 1 Proof#1 Since x...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online