Questions: Sequences: Epsilon-N Convergence

Question 1

Given ε > 0, which choice of N guarantees that |1/n − 0| < ε for all n > N?

• A. N = ε
• B. N = 1/ε
• C. N = ε²
• D. N = √ε

Answer: (B)

We need 1/n < ε, which rearranges to n > 1/ε. Choosing N = 1/ε (or ⌈1/ε⌉ for an integer) ensures that any n > N satisfies n > 1/ε and hence 1/n < ε. The other choices do not correctly solve the inequality 1/n < ε.

Question 2

If a sequence (aₙ) converges to L, then nearly every term of the sequence is expected to satisfy |aₙ − L| < ε for any ε > 0.

• T. True
• F. False

Answer: False

Convergence only requires that terms eventually stay within ε of L — meaning for all n beyond some cutoff N. Finitely many early terms (before N) can be arbitrarily far from L without violating the definition. For example, the sequence (100, 200, 1/3, 1/4, 1/5, …) converges to 0 even though its first two terms are far away.

Question 3

In the epsilon-N definition, why must N be found for *every* ε > 0, including arbitrarily small values, rather than just for one specific small ε?

Model answer: Finding N for a single ε only shows the sequence gets close once — it does not rule out the sequence drifting away and then approaching L again for smaller ε. The universal quantifier 'for every ε' ensures the sequence stays arbitrarily close to L permanently and completely, not just at one tolerance level.

Convergence means the sequence eventually gets as close to L as anyone could demand. If we only verified one specific ε, the sequence might happen to satisfy that tolerance while failing a stricter one. The quantifier 'for every ε' is what makes the limit unique and the convergence unconditional — it is the formal content of 'arbitrarily close.'