Unit III: Real Analysis - Sequences

1. Subsequences

A fundamental concept in real analysis is the behavior of a sequence when we only look at a selected infinite subset of its terms. This behavior is captured by subsequences.

Definition

Let (x_n) be a sequence of real numbers and let n₁ < n₂ < ... < n_k < ... be a strictly increasing sequence of natural numbers. The sequence (x_nk) given by (x_n1, x_n2, x_n3, ...) is called a subsequence of (x_n).

Crucial Property of Indexing

Because the sequence of indices (n_k) consists of strictly increasing natural numbers, we have an important inequality that is used in almost every proof involving subsequences:

n_k ≥ k for all k ∈ N

Examples of Subsequences

• If (x_n) = (1, 1/2, 1/3, 1/4, ...), then (x_2k) = (1/2, 1/4, 1/6, ...) is the subsequence of even terms.
• If (x_n) = ((-1)ⁿ), then (x_2k) = (1, 1, 1, ...) is a constant subsequence, and (x_2k-1) = (-1, -1, -1, ...) is another constant subsequence.

Theorem: Convergence of Subsequences

If a sequence of real numbers (x_n) converges to a real number x, then any subsequence (x_nk) of (x_n) also converges to x.

Step-by-Step Proof:

1. Suppose lim (x_n) = x. This means for any ε > 0, there exists a natural number K such that for all n ≥ K, |x_n - x| < ε.
2. Let (x_nk) be a subsequence of (x_n). Since (n_k) is strictly increasing, we have n_k ≥ k for all k.
3. If we choose k ≥ K, then n_k ≥ k ≥ K.
4. Consequently, for all k ≥ K, the inequality |x_nk - x| < ε holds.
5. This proves that lim (x_nk) = x.

2. Divergence Criteria

We can use subsequences as a powerful tool to prove that a sequence does not converge. This is formalized in the divergence criteria.

The Divergence Theorems

Divergence Criterion 1: If a sequence (x_n) has two subsequences (x_nk) and (x_mk) that converge to different limits L₁ ≠ L₂, then the sequence (x_n) is divergent.

Divergence Criterion 2: If a sequence (x_n) has a subsequence (x_nk) that is divergent, then the sequence (x_n) is divergent.

Practical Example

Prove that the sequence x_n = (-1)ⁿ diverges.

Step-by-step Solution:

1. Consider the subsequence of even terms: x_2k = (-1)^2k = 1. This constant subsequence converges to 1.
2. Consider the subsequence of odd terms: x_2k-1 = (-1)^2k-1 = -1. This constant subsequence converges to -1.
3. Since we have found two subsequences that converge to different limits (1 ≠ -1), the divergence criterion guarantees that the original sequence (x_n) is divergent.

3. Monotone Subsequence Theorem

Not all sequences are monotone (increasing or decreasing), but amazingly, every sequence contains a monotone portion inside it.

Theorem Statement

Monotone Subsequence Theorem: Every sequence of real numbers has a monotone subsequence.

The Concept of a "Peak"

To prove this theorem, we define a special type of term called a peak.

A term x_m in a sequence (x_n) is called a peak if x_m ≥ x_n for all n > m.

In simple terms, a peak is a term that is greater than or equal to all subsequent terms in the sequence.

Proof Outline

Let (x_n) be any sequence of real numbers. There are two mutually exclusive cases regarding the number of peaks in the sequence:

Detailed Analysis of Case 1 (Infinitely Many Peaks):

If there are infinitely many peaks, we can list them in order of their occurrence: x_m1, x_m2, x_m3, ... where m₁ < m₂ < m₃ < ...

Since x_m1 is a peak and m₂ > m₁, we must have x_m1 ≥ x_m2.

Since x_m2 is a peak and m₃ > m₂, we must have x_m2 ≥ x_m3.

Continuing this process, we obtain a decreasing subsequence: x_m1 ≥ x_m2 ≥ x_m3 ≥ ...

Detailed Analysis of Case 2 (Finitely Many Peaks):

If there are finitely many peaks, let there be a natural number S such that there are no peaks at or after index S. (If there are no peaks at all, we can set S = 1).

1. Let n₁ = S. Since n₁ ≥ S, x_n1 is not a peak.
2. By definition of "not a peak", there must exist some index n₂ > n₁ such that x_n2 > x_n1.
3. Since n₂ > n₁ ≥ S, x_n2 is also not a peak.
4. Therefore, there must exist some index n₃ > n₂ such that x_n3 > x_n2.
5. Repeating this step indefinitely, we construct a strictly increasing subsequence: x_n1 < x_n2 < x_n3 < ...

4. Bolzano-Weierstrass Theorem

The Bolzano-Weierstrass Theorem establishes a deep connection between boundedness and convergence.

Theorem Statement

Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a convergent subsequence.

Proof Using Monotone Subsequence Theorem

This theorem can be elegantly proven using our previous results:

1. Let (x_n) be a bounded sequence of real numbers. This means there exists a real number M > 0 such that |x_n| ≤ M for all n.
2. By the Monotone Subsequence Theorem, the sequence (x_n) must contain a monotone subsequence (x_nk).
3. Since the original sequence (x_n) is bounded, the subsequence (x_nk) must also be bounded (since |x_nk| ≤ M for all k).
4. According to the Monotone Convergence Theorem, any sequence that is both monotone and bounded must converge.
5. Therefore, the monotone subsequence (x_nk) is convergent.

Important Observation: Even if a sequence itself does not converge (such as x_n = (-1)ⁿ), as long as it is bounded, we can always extract a convergent subsequence (e.g., the constant subsequence x_2k = 1).

5. Cauchy Sequences

In many situations, we want to know if a sequence converges, but finding the exact limit first is extremely difficult. A Cauchy sequence allows us to analyze convergence purely by looking at the relationships between the terms themselves.

Definition

A sequence (x_n) of real numbers is called a Cauchy sequence if for every ε > 0, there exists a natural number H such that for all n, m ≥ H, the distance between the terms satisfies:
|x_n - x_m| < ε

Intuitive Concept

In a convergent sequence, the terms get closer and closer to a fixed target limit. In a Cauchy sequence, the terms simply get closer and closer to each other.

Theorem: Convergent implies Cauchy

Every convergent sequence of real numbers is a Cauchy sequence.

Proof:

1. Suppose (x_n) converges to x.
2. Let ε > 0. Since lim x_n = x, there exists a natural number K such that for all n ≥ K, |x_n - x| < ε/2.
3. Let n, m ≥ K. By the triangle inequality:
   |x_n - x_m| = |(x_n - x) + (x - x_m)| ≤ |x_n - x| + |x_m - x|
4. Since both n, m ≥ K, we have:
   |x_n - x_m| < ε/2 + ε/2 = ε
5. Therefore, (x_n) is a Cauchy sequence.

Lemma: Cauchy Sequences are Bounded

Every Cauchy sequence of real numbers is bounded.

Step-by-Step Proof:

1. Let (x_n) be a Cauchy sequence. Let ε = 1.
2. By definition, there exists a natural number H such that for all n, m ≥ H, |x_n - x_m| < 1.
3. In particular, if we fix m = H, then for all n ≥ H, we have |x_n - x_H| < 1.
4. By the reverse triangle inequality, this implies |x_n| < |x_H| + 1 for all n ≥ H.
5. To find a bound that works for the entire sequence (including the terms before H), we define:
   M = max{|x₁|, |x₂|, ..., |x_H-1|, |x_H| + 1}
6. It follows that |x_n| ≤ M for all n ∈ N. Thus, the sequence is bounded.

6. Cauchy Convergence Criterion

The culmination of these concepts is the Cauchy Convergence Criterion, which asserts that in the real number system, being a Cauchy sequence is exactly equivalent to being a convergent sequence.

Theorem Statement

Cauchy Convergence Criterion: A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof of the Criterion

We already proved that "Convergent → Cauchy". Now we must prove the reverse: "Cauchy → Convergent".

1. Let (x_n) be a Cauchy sequence.
2. We know that every Cauchy sequence is bounded. Thus, (x_n) is bounded.
3. By the Bolzano-Weierstrass Theorem, the bounded sequence (x_n) has a convergent subsequence (x_nk) that converges to some real limit x*.
4. We will show that the entire sequence (x_n) converges to x*.
5. Let ε > 0. Since (x_n) is Cauchy, there exists a natural number H such that for all n, m ≥ H:
   |x_n - x_m| < ε/2
6. Since the subsequence (x_nk) converges to x*, there exists a term in this subsequence, say x_nK, such that n_K ≥ H and:
   |x_nK - x*| < ε/2
7. Now, let n ≥ H. We can write:
   |x_n - x*| = |(x_n - x_nK) + (x_nK - x*)| ≤ |x_n - x_nK| + |x_nK - x*|
8. Since n ≥ H and n_K ≥ H, we have |x_n - x_nK| < ε/2. Combining this with |x_nK - x*| < ε/2, we get:
   |x_n - x*| < ε/2 + ε/2 = ε
9. This proves that the sequence (x_n) converges to x*.

Practical Application: Testing for Convergence

Show that the sequence x_n = 1 + 1/2! + 1/3! + ... + 1/n! converges.

Proof using Cauchy Criterion:

1. Let m > n. Let's look at the difference:
   |x_m - x_n| = 1/(n+1)! + 1/(n+2)! + ... + 1/m!
2. Since k! ≥ 2^k-1 for all k, we have:
   |x_m - x_n| ≤ 1/2ⁿ + 1/2ⁿ⁺¹ + ... + 1/2^m-1
3. This is a geometric series sum:
   1/2ⁿ (1 + 1/2 + ... + 1/2^m-n-1) < 1/2ⁿ × 2 = 1/2^n-1
4. For any ε > 0, we can choose H such that 1/2^H-1 < ε. Then for all m > n ≥ H:
   |x_m - x_n| < 1/2^n-1 ≤ 1/2^H-1 < ε
5. This proves that (x_n) is a Cauchy sequence, and by the Cauchy Criterion, it must converge.

7. Properly Divergent Sequences

Not all divergent sequences behave the same way. Some oscillate, while others grow without bound in a specific direction. The latter are called properly divergent.

Definitions

A sequence (x_n) of real numbers is said to diverge to +∞ if for every real number α, there exists a natural number K such that for all n ≥ K,
x_n > α
We write this as lim x_n = +∞.

A sequence (x_n) of real numbers is said to diverge to -∞ if for every real number β, there exists a natural number K such that for all n ≥ K,
x_n < β
We write this as lim x_n = -∞.

A sequence is properly divergent if it either diverges to +∞ or diverges to -∞.

Properties of Properly Divergent Sequences

• A monotone sequence is properly divergent if and only if it is unbounded.
• Comparison Theorem: If (x_n) and (y_n) are sequences such that x_n ≤ y_n for all n, then:
  • If lim x_n = +∞, then lim y_n = +∞.
  • If lim y_n = -∞, then lim x_n = -∞.

Summary Comparison Table

Common Academic Mistakes to Avoid

• Mistake: Assuming that if a sequence has a convergent subsequence, then the sequence itself must converge.
  Correction: The sequence x_n = (-1)ⁿ has the convergent subsequence x_2k = 1, but the main sequence diverges.
• Mistake: Thinking that "divergent" is the same as "properly divergent".
  Correction: The sequence (-1)ⁿ is divergent, but it is not properly divergent because it does not trend to +∞ or -∞.
• Mistake: Trying to apply the Cauchy criterion by showing |x_n+1 - x_n| → 0.
  Correction: Showing that consecutive terms get closer is not enough! For example, if x_n = 1 + 1/2 + 1/3 + ... + 1/n (the harmonic series), we have |x_n+1 - x_n| = 1/(n+1) → 0, but the sequence diverges to +∞ and is not Cauchy. You must show |x_m - x_n| < ε for all m, n ≥ H.