Real Analysis 2 | Sequences and Limits [dark version]

Q: How does defining a sequence as a function clarify what a_n means?

A sequence is treated as a map from the natural numbers to the real numbers. That means each natural number n is an input, and the output is the real number a_n. Writing a_n in parentheses (a_n) emphasizes it is an infinite list, and the indexing convention (starting at 1 or at 0) just changes where the list begins, not the underlying functional idea.

Q: Why does a_n = (-1)^n fail to converge?

The terms alternate forever: a_1 = -1, a_2 = 1, a_3 = -1, and so on. No matter how large n gets, the sequence keeps jumping between -1 and 1, so there is no single real number a such that all sufficiently late terms stay within ε of a for every ε > 0.

Q: What does the ε-neighborhood condition require exactly?

For convergence to a, the transcript uses the interval (a − ε, a + ε), described as the ε-neighborhood of a. The formal requirement is: there exists N such that for every n ≥ N, the distance from a_n to a is less than ε, written as |a_n − a| < ε. Only finitely many terms before N may violate the condition.

Q: How is the proof for 1/n → 0 structured?

It begins with an arbitrary ε > 0 (since convergence must work for every tolerance). Then it rewrites the distance as |a_n − 0| = 1/n. The proof chooses N large enough that 1/n 1, which ensures 1/N < ε and therefore 1/n ≤ 1/N < ε for later indices.

TL;DR

A real sequence can be defined as a function from natural numbers to real numbers, with a_n as the output for input n.

Briefing Cornell Notes

Briefing

Sequences are defined as functions from the natural numbers into the real numbers, turning an index like n into a specific real value a_n. That definition matters because it lets limits be treated with precise logic: instead of saying numbers “settle down,” convergence is framed as a property that must hold for every tolerance you choose. The transcript emphasizes that the indexing choice (starting at 1 or at 0) is mostly a convention, while the underlying idea remains the same—each natural number picks out one term of the sequence.

To build intuition, several sequences are used to show how behavior changes as n grows. For a_n = (-1)^n, the terms alternate forever between -1 and 1, so the values never get closer to a single number. Visualizing such sequences on a number line highlights the key issue: even as n increases without bound, the sequence keeps “jumping” between two fixed values. Another example, a_n = 1/n, produces terms 1, 1/2, 1/3, 1/4, …, which steadily approach 0. A third example, powers of 2 (2, 4, 8, 16, …), grows without an upper bound, motivating the need to clarify what “limit” could mean in cases where values diverge to infinity.

The core formal definition introduced is convergence. A sequence a_n is said to converge to a real number a if, for every positive tolerance ε, all terms from some point onward lie within ε of a. The “ε-neighborhood” is described as the interval (a − ε, a + ε), and the condition is written using absolute value: there must exist an integer N such that for every n ≥ N, |a_n − a| < ε. The transcript stresses the “eventually” part: only finitely many early terms may fall outside the neighborhood; after N, the sequence must stay inside it.

If no such number a exists that satisfies the condition for every ε, the sequence is called divergent. The transcript then demonstrates how to prove convergence using the example 1/n → 0. The proof starts by taking an arbitrary ε > 0, then rewriting the distance as |a_n − 0| = 1/n. The task becomes ensuring 1/n < ε for all sufficiently large n. This is achieved by choosing N large enough so that n ≥ N implies 1/n ≤ 1/N < ε, which is arranged by selecting N so that Nε > 1. The argument relies on the basic arithmetic fact that multiplying a small ε by a sufficiently large integer can exceed any fixed number.

Overall, the lesson is both conceptual and procedural: convergence is not a vague trend but a universal “for every ε” requirement, and proofs follow a predictable structure—start with what must be shown, pick an arbitrary tolerance, and then choose an index N that forces all later terms to stay within that tolerance.

Cornell Notes

A sequence is a function a: N → R, where each index n produces a real term a_n. Convergence is defined using ε-neighborhoods: a_n converges to a if for every ε > 0 there exists an integer N such that for all n ≥ N, |a_n − a| < ε. The “eventually” clause is essential—only finitely many early terms may lie outside the interval (a − ε, a + ε). Using examples, (-1)^n does not converge because it keeps alternating between -1 and 1, while 1/n converges to 0 because its terms get arbitrarily close to zero. Divergence means no real number a satisfies the ε-based condition.

How does defining a sequence as a function clarify what a_n means?

A sequence is treated as a map from the natural numbers to the real numbers. That means each natural number n is an input, and the output is the real number a_n. Writing a_n in parentheses (a_n) emphasizes it is an infinite list, and the indexing convention (starting at 1 or at 0) just changes where the list begins, not the underlying functional idea.

Why does a_n = (-1)^n fail to converge?

The terms alternate forever: a_1 = -1, a_2 = 1, a_3 = -1, and so on. No matter how large n gets, the sequence keeps jumping between -1 and 1, so there is no single real number a such that all sufficiently late terms stay within ε of a for every ε > 0.

What makes the sequence a_n = 1/n converge, and to what limit?

As n increases, 1/n produces 1, 1/2, 1/3, 1/4, …, getting closer and closer to 0. Formally, for any ε > 0 one can choose N so that for all n ≥ N, 1/n < ε. Since |1/n − 0| = 1/n, the ε-condition is satisfied with limit a = 0.

What does the ε-neighborhood condition require exactly?

For convergence to a, the transcript uses the interval (a − ε, a + ε), described as the ε-neighborhood of a. The formal requirement is: there exists N such that for every n ≥ N, the distance from a_n to a is less than ε, written as |a_n − a| < ε. Only finitely many terms before N may violate the condition.

How is the proof for 1/n → 0 structured?

It begins with an arbitrary ε > 0 (since convergence must work for every tolerance). Then it rewrites the distance as |a_n − 0| = 1/n. The proof chooses N large enough that 1/n < ε for all n ≥ N; one way is to pick N so that Nε > 1, which ensures 1/N < ε and therefore 1/n ≤ 1/N < ε for later indices.

Review Questions

State the formal definition of convergence of a real sequence using ε and N.
Give an example of a sequence that does not converge and explain which part of the ε-condition fails.
Outline the key steps needed to prove that a_n = 1/n converges to 0.

Key Points

1
A real sequence can be defined as a function from natural numbers to real numbers, with a_n as the output for input n.
2
Convergence is defined using ε-neighborhoods: for every ε > 0, all terms from some index onward must lie within ε of the proposed limit.
3
The “eventually” requirement means only finitely many early terms may be outside the ε-neighborhood; later terms must stay inside.
4
The sequence (-1)^n diverges because it keeps alternating between -1 and 1 and never settles near a single real number.
5
The sequence 1/n converges to 0 because 1/n can be made smaller than any ε by choosing n large enough.
6
A standard convergence proof starts by fixing an arbitrary ε > 0, then selecting N to force |a_n − a| < ε for all n ≥ N.

Highlights

Convergence is a universal condition: it must hold for every ε > 0, not just for a convenient tolerance.

The “eventually” clause is the heart of the definition—after some N, the sequence must stay within the ε-neighborhood forever.

For 1/n → 0, the distance simplifies to |1/n − 0| = 1/n, and choosing N so that Nε > 1 makes the inequality work.

(-1)^n never converges because its terms keep oscillating between two values regardless of how large n becomes.

Topics

Sequences
Limits
Convergence
Divergence
ε-Definitions