# Sequence limits and epsilon proofs

> For every ε>0 there exists N such that every n≥N satisfies the error bound.

Course: [Real analysis: limits and rigorous reasoning](https://studybrix.com/en/courses/real-analysis/index.md)
Level: advanced
Prerequisites: Logic, inequalities and real sequences.

## Notes

### Quantifier order

For every ε>0 there exists N such that every n≥N satisfies the error bound.

### Construct N

Solve the desired inequality for n, then choose an integer cutoff meeting it.

### Cauchy criterion

A real sequence converges exactly when it is Cauchy, using completeness of the reals.

### Counterexamples

One carefully verified counterexample refutes a universal claim; many examples do not prove it.

## Practice questions

### 1. Prove 1/n tends to zero using the epsilon definition.

Sample answer / marking guidance: Given ε>0, choose an integer N>1/ε. For n≥N, 0<1/n≤1/N<ε.

Explanation: Credit the order of quantifiers, an explicit N, and a bound for every n≥N.

### 2. Prove every convergent real sequence is bounded.

Sample answer / marking guidance: If a_n→L, eventually abs(a_n−L)<1, so the tail is bounded by abs(L)+1. Take the maximum of that bound and the finitely many initial absolute values.

Explanation: Credit both tail control and the finite initial segment.

### 3. Which statement reverses the quantifiers incorrectly for convergence?

A. There exists one N that works for every ε>0
B. All sufficiently late terms lie within ε
C. The cutoff may depend on ε
D. For each ε>0 there is a suitable N

Answer: There exists one N that works for every ε>0

Explanation: The needed cutoff generally grows as ε shrinks.

### 4. What is the smallest positive integer N such that 1/n<0.01 for every n≥N?

Answer: 101

Explanation: The strict inequality requires n>100, not n≥100.

### 5. Give a bounded real sequence that does not converge and justify it.

Sample answer / marking guidance: a_n=(−1)^n is bounded by 1 in absolute value, but its even and odd subsequences have limits 1 and −1.

Explanation: Credit boundedness and a valid nonconvergence argument.

### 6. What extra property makes every Cauchy sequence converge in R?

A. Completeness
B. Countability
C. A chosen decimal display precision
D. All terms being integers

Answer: Completeness

Explanation: Completeness prevents a Cauchy sequence from approaching a missing real-number limit.

### 7. Prove a real sequence cannot converge to two different limits L and M.

Sample answer / marking guidance: Assume L≠M and choose ε=abs(L−M)/3. Far enough out, the triangle inequality would give abs(L−M)<2ε, a contradiction.

Explanation: Credit one common tail index and a contradiction via the triangle inequality.

### 8. Does checking the first million terms establish convergence of every sequence?

A. No
B. Only when a computer checks them
C. Yes, by definition
D. Yes if the values are positive

Answer: No

Explanation: Convergence is a statement about all sufficiently large indices, not a finite prefix.

## Further reading

- https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/
- https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.