University:
The University Of Western Ontario
Course:
MATH 1000A | Introduction to Mathematical Analysis
Academic year:

2024
Views:

147

Pages:

3
Author:

Kyuwon J.

Limit of a Sequence Definition of the Limit of a Sequence A sequence {an } is said to converge to a limit L ∈ R if, for every ε > 0, there exists N ∈ N such that for all n ≥ N , we have: |an − L| < ε. This definition means that the terms of the sequence {an } can get arbitrarily close to L as n → ∞. In this case, L is called the limit of the sequence, and we write: lim an = L or simply an → L as n → ∞. n→∞ — Practice Problems and Step-by-Step Solutions Problem 1: Prove that the sequence {an } = Step-by-Step Solution: n n+1 converges to 1. 1. Start with the definition of the limit. We want to show that for any ε > 0, there exists N ∈ N such that for all n ≥ N : |an − 1| < ε. 2. Compute |an − 1|: |an − 1| = 3. To make 1 n+1 −1 1 n = −1 = . n+1 n+1 n+1 < ε, choose N > 1ε . For all n ≥ N , we have n + 1 ≥ N + 1 > 1ε , so: 1 < ε. n+1 4. Therefore, |an − 1| < ε for all n ≥ N . This proves that: lim an = 1. n→∞ — Problem 2: Show that the sequence {an } = Step-by-Step Solution: √1 n converges to 0. 1. Start with the definition of the limit. We want to show that for any ε > 0, there exists N ∈ N such that for all n ≥ N : |an − 0| = |an | < ε. 1 2. Compute |an |: 1 |an | = √ . n 3. To make √1 n < ε, choose N > √ 1 . ε2 n> For all n ≥ N , we have n ≥ N > 1 ε and hence 1 , ε2 so: 1 √ < ε. n 4. Therefore, |an − 0| < ε for all n ≥ N . This proves that: lim an = 0. n→∞ — Problem 3: Does the sequence {an } = (−1)n + Step-by-Step Solution: 1 n converge? 1. Split the sequence into two components: {(−1)n } and { n1 }. Analyze each component separately. 2. The term (−1)n alternates between 1 and −1, so it does not converge to any single value. 3. The term n1 converges to 0 as n → ∞ because for any ε > 0, we can choose N > such that for all n ≥ N : 1 < ε. n 1 ε 4. However, the presence of the non-convergent (−1)n term prevents the entire sequence from converging to a single value. 5. Therefore, the sequence {an } diverges. — Problem 4: Prove that the constant sequence {an } = c converges to c. Step-by-Step Solution: 1. Start with the definition of the limit. We want to show that for any ε > 0, there exists N ∈ N such that for all n ≥ N : |an − c| < ε. 2. Compute |an − c|: |an − c| = |c − c| = 0. 3. Since |an − c| = 0 for all n, the inequality |an − c| < ε holds for any choice of N ∈ N. 4. Thus, limn→∞ an = c. 2 — Problem 5: Show that {an } = Step-by-Step Solution: 2n+3 n+5 converges to 2. an = 2n + 3 −7 =2+ . n+5 n+5 1. Simplify the sequence: −7 approaches 0 as n → ∞ because for any ε > 0, we can choose N > 7ε − 5 2. The term n+5 such that for all n ≥ N : −7 < ε. n+5 3. Thus, as n → ∞, an = 2 + −7 n+5 → 2. 4. Therefore, limn→∞ an = 2. 3