Phase 1 - Lesson 1.5

Sequences, Limits, and Reading & Writing Proofs

The epsilon-N definition of a limit - your first genuinely rigorous definition - and habits for reading and writing proofs that others can check.

⏱ 55 min● Intermediate🔗 Prereqs: 1.4
↖ Phase 1 hub
Builds on: 1.2 gave nested quantifiers; the limit definition is the payoff - a ∀∃∀ statement you now can parse.
Leads to: This epsilon-N template becomes epsilon-delta continuity (Phase 2) and convergence in probability (Phase 7).

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Key Vocabulary

Sequence
A function from the naturals to the reals, written (a_n); the ordered list a_1,a_2,….
Convergence
(a_n) converges to L if its terms are eventually arbitrarily close to L; written a_n → L.
Epsilon-N definition
∀ε>0 ∃N such that n≥N ⇒ |a_n − L| < ε.
Bounded sequence
There is M with |a_n| ≤ M for all n.
Monotone sequence
Non-increasing or non-decreasing for all n.
Divergence
Failure to converge to any finite L; e.g. oscillation or growth without bound.

What a limit really says

Intuitively \(a_n\to L\) means ‘the terms get and stay close to \(L\)’. Rigor makes ‘close’ and ‘stay’ precise with the definition you spent 1.2 preparing for:

\[a_n\to L\ \iff\ \forall \varepsilon\gt 0\ \exists N\in\N\ \text{s.t.}\ n\ge N\ \Rightarrow\ |a_n-L|\lt \varepsilon\] (1.5)

Read it as a game: an adversary picks a tolerance \(\varepsilon\); you must respond with a threshold \(N\) past which every term is within \(\varepsilon\) of \(L\). You win for every \(\varepsilon\) exactly when the limit holds. Smaller \(\varepsilon\) typically forces larger \(N\).

Worked Example - Prove 1/n → 0
1
Fix \(\varepsilon\gt 0\). We need \(N\) with \(n\ge N\Rightarrow |1/n-0|\lt \varepsilon\), i.e. \(1/n\lt \varepsilon\).
2
That is \(n\gt 1/\varepsilon\). So choose any integer \(N\gt 1/\varepsilon\) (the Archimedean property guarantees one exists).
3
Then for \(n\ge N\): \(1/n\le 1/N\lt \varepsilon\). Since \(\varepsilon\) was arbitrary, \(1/n\to0\). □

Monotone convergence: a limit without computing it

You can know a limit exists without a formula for it. The Monotone Convergence Theorem: a bounded, monotone sequence converges (to its supremum if increasing, its infimum if decreasing). This is why \((1+1/n)^n\) converges - it is increasing and bounded above - and its limit is named \(e\).

Intuition
A rising sequence with a ceiling has nowhere to go but toward a specific height. Existence of that height is guaranteed by the completeness of the real numbers - the property that distinguishes ℝ from the rationals.

Reading and writing proofs

Reading a proof is active work: for each line ask ‘what justifies this - a definition, a hypothesis, or a previous line?’. Writing a proof is the mirror: make every justification explicit, define notation before using it, and state clearly where each hypothesis is used.

Reader's questionWhat a good proof supplies
Why is this step allowed?An explicit definition, theorem, or prior line
Where is hypothesis H used?A visible appeal to H at the right moment
What is this symbol?Introduced and typed before first use
Is the ‘arbitrary’ element really arbitrary?No hidden special assumptions about it

Interactive: watch terms fall inside the epsilon band

Common Mistakes to Avoid
  • Letting N depend on n. N may depend on ε only; it is chosen before n ranges past it.
  • Proving |a_n − L| < ε for one convenient ε instead of every ε > 0.
  • Assuming a bounded sequence converges - it need not (e.g. (−1)^n); monotonicity is also required.
  • Writing ‘clearly’ or ‘obviously’ to paper over the one step you did not actually check.
Quant Practitioner Tips
  • To prove a limit, work backwards from |a_n − L| < ε to an inequality on n; that reveals the N to pick.
  • State ‘let ε > 0 be arbitrary’ first and ‘since ε was arbitrary’ last - the frame of every epsilon proof.
  • When no formula for the limit is available, look for monotone + bounded and invoke Monotone Convergence.
  • Read published proofs with a pen: reconstruct each skipped step yourself. That is retrieval practice for proof-writing.

Knowledge Check

Q1 Medium
In the definition ‘∀ε>0 ∃N: n≥N ⇒ |a_n−L|<ε’, the threshold N is allowed to depend on:
n
ε only
both n and ε
nothing
Q2 Medium
Which sequence converges by the Monotone Convergence Theorem?
a_n=(−1)^n
a_n=n
a_n=1−1/n
a_n=sin n
Q3 Hard
To prove 1/n² → 0, choosing N works if for n≥N we get 1/n²<ε. A valid choice of N is any integer greater than:
ε
1/ε
1/√ε
√ε

Practical Exercise

Using the epsilon-N definition, prove that \(a_n=\dfrac{2n+1}{n}\) converges, and identify the limit.

▶ Show full solution

Guess the limit: \(a_n=2+\tfrac1n\to 2\). Prove it. Fix \(\varepsilon\gt 0\). Then

\[|a_n-2|=\Big|\,\tfrac{2n+1}{n}-2\,\Big|=\Big|\tfrac1n\Big|=\tfrac1n.\]

We need \(1/n\lt \varepsilon\), i.e. \(n\gt 1/\varepsilon\). Choose any integer \(N\gt 1/\varepsilon\). Then for \(n\ge N\), \(|a_n-2|=1/n\le 1/N\lt \varepsilon\). Since \(\varepsilon\) was arbitrary, \(a_n\to 2\). □

After the reveal, answer for yourself: The algebra collapsed |a_n − L| to 1/n before choosing N. Simplifying the distance first almost always exposes the right N.

Lesson Summary

A sequence converges to L exactly when, for every tolerance ε, all terms past some index N stay within ε of L - the ∀ε∃N template you now parse and produce. Some limits are proved to exist without a formula via Monotone Convergence (bounded + monotone). Reading proofs actively and writing them with every justification explicit is the craft this phase has been building toward.

Formula Sheet Additions

Epsilon-N limit
\[a_n\to L\iff\forall\varepsilon\gt 0\,\exists N:\,n\ge N\Rightarrow|a_n-L|\lt \varepsilon\]
The template that becomes epsilon-delta continuity and stochastic convergence later.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: In the epsilon-N definition, what may N depend on, and why does that matter?
A: N may depend only on ε (chosen before n is quantified). If N could depend on n the definition would be trivially satisfiable and meaningless.
Q: State the Monotone Convergence Theorem and why it is useful.
A: A bounded monotone real sequence converges (increasing → its supremum, decreasing → its infimum); it proves existence of a limit even when no closed-form value is available, e.g. defining e.

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Source References

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