Sequences and Convergence
TertiaryReal Analysis
Concept
Convergence of sequences, Cauchy sequences, and the Bolzano-Weierstrass theorem.
Walkthrough
Using the ε-N definition, prove that the sequence aₙ = 1/n converges to 0.
- 1.Recall the ε-N definition: aₙ → L if for every ε > 0, there exists N such that |aₙ − L| < ε for all n ≥ N.
- 2.We need to show: for every ε > 0, there exists N such that |1/n − 0| = 1/n < ε for all n ≥ N.
- 3.Solve 1/n < ε: this is equivalent to n > 1/ε.
- 4.Choose N = ⌈1/ε⌉ (the smallest integer ≥ 1/ε). Then for all n ≥ N, we have n ≥ N ≥ 1/ε, so 1/n ≤ 1/N < ε. ✓
- 5.Since such an N exists for every ε > 0, the sequence converges to 0.
Answer
Converges to 0
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