hw2sol

# hw2sol - Math 3110 Homework 2 Solutions Exercise 3.1.2...

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

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

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

Unformatted text preview: Math 3110 Homework 2 Solutions Exercise 3.1.2 Claim. If a n is a non-negative sequence, lim a n = 0 = ⇒ lim √ a n = 0 . Proof. Let a n be a non-negative sequence. Assume lim a n = 0 (this means that for any δ > 0, there is a natural number N such that for all n ≥ N , | a n | < δ ). Let > 0 and set δ = 2 . Since lim a n = 0, then there exists a natural number N such that for all n ≥ N , | a n | < δ = 2 , which implies that √ a n < . That for every > 0, such an N exists, means that lim √ a n = 0. Exercise 3.2.5 Let { a n } be a convergent sequence of integers, with a n → L . We prove that { a n } is eventually constant, that is, a n = L for large n . Since a n → L , then there is some natural number some N such that | a n- L | < 1 4 for all n ≥ N . This requires a n to lie in the interval ( L- 1 4 ,L + 1 4 ) for all n ≥ N . As this interval has length 1 2 , it can contain at most 1 2 = 1 integer, say, M . Thus, a n = M for all n ≥ N , which is to say, { a n } is eventually constant.is eventually constant....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

hw2sol - Math 3110 Homework 2 Solutions Exercise 3.1.2...

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

View Full Document
Ask a homework question - tutors are online