Unit II: Sequence of Real Numbers

1. Sequence of Real Numbers

A sequence of real numbers is a foundational concept in mathematical analysis that extends the idea of an ordered list of infinite numbers.

Formal Definition: A sequence of real numbers is a function f: ℕ → ℝ, where the domain is the set of natural numbers ℕ = {1, 2, 3, ...} and the codomain is the set of real numbers ℝ.

If f(n) = a_n for n ∈ ℕ, we write the sequence as (a_n), {a_n}, or (a₁, a₂, a₃, ...). The value a_n is referred to as the n-th term or the general term of the sequence.

Key Examples of Sequences

• Constant Sequence: a_n = c for some c ∈ ℝ. The sequence is (c, c, c, ...).
• Arithmetic Progression Sequence: a_n = a + (n - 1)d. For example, if a = 1, d = 2, we have (1, 3, 5, 7, ...).
• Geometric Sequence: a_n = rⁿ. For r = 1/2, we have (1/2, 1/4, 1/8, ...).
• Alternating Sequence: a_n = (-1)ⁿ. The sequence is (-1, 1, -1, 1, ...).

2. Limit of a Sequence

The limit of a sequence describes the ultimate behavior of the terms a_n as n grows infinitely large (n → ∞).

Definition (ε-N Convergence): A sequence (a_n) is said to converge to a real number L if, for every ε > 0, there exists a natural number N ∈ ℕ (which generally depends on ε) such that for all n ≥ N, the inequality |a_n - L| < ε holds.

If (a_n) converges to L, we write:

lim_n→∞ a_n = L   or   a_n → L as n → ∞

The number L is called the limit of the sequence. If no such real number L exists, the sequence is said to diverge.

Geometrical Interpretation

Geometrically, the limit definition means that if we place an open interval (L - ε, L + ε) around the limit L, then only a finite number of terms of the sequence (namely a₁, a₂, ..., a_N-1) can lie outside this interval. All subsequent terms from a_N onwards must lie inside the interval.

Step-by-Step Proof Example

Problem: Prove using the formal definition that lim_n→∞ (1 / n) = 0.

1. Let ε > 0 be given.
2. We must find an N ∈ ℕ such that for all n ≥ N, |(1 / n) - 0| < ε.
3. Simplify the expression: |1 / n| = 1 / n. So we need 1 / n < ε, which is equivalent to n > 1 / ε.
4. By the Archimedean Property of real numbers, there exists a natural number N such that N > 1 / ε.
5. Thus, for all n ≥ N, we have n > 1 / ε, which implies 1 / n < ε.
6. Therefore, |(1 / n) - 0| < ε for all n ≥ N, completing the proof.

3. Uniqueness of Limits

A sequence cannot settle on two different target values simultaneously. This is formalised by the uniqueness theorem.

Theorem: If a sequence (a_n) of real numbers converges, its limit is unique.

Proof: We prove this by contradiction.

1. Assume that (a_n) converges to two distinct limits, L₁ and L₂, with L₁ ≠ L₂.
2. This implies that the distance |L₁ - L₂| > 0. Let us choose ε = |L₁ - L₂| / 2.
3. Since lim_n→∞ a_n = L₁, there exists N₁ ∈ ℕ such that for all n ≥ N₁, |a_n - L₁| < ε.
4. Since lim_n→∞ a_n = L₂, there exists N₂ ∈ ℕ such that for all n ≥ N₂, |a_n - L₂| < ε.
5. Let N = max(N₁, N₂). For all n ≥ N, both inequalities hold simultaneously.
6. Applying the triangle inequality:
|L₁ - L₂| = |(L₁ - a_n) + (a_n - L₂)| ≤ |L₁ - a_n| + |a_n - L₂| < ε + ε = 2ε = |L₁ - L₂|.
7. This yields the strict inequality |L₁ - L₂| < |L₁ - L₂|, which is a contradiction.
8. Hence, our assumption was false, and we must have L₁ = L₂.

4. Recursive Sequences

A recursive sequence (or inductively defined sequence) is defined by giving one or more initial terms, and defining subsequent terms as a function of prior terms.

Definition: A sequence (a_n) is recursive if a₁ is specified, and a_n+1 = f(a_n) for all n ∈ ℕ, where f is a real-valued function.

Example: Let a₁ = 1, and let a_n+1 = 3a_n + 1. The terms of this sequence are generated as: a₁ = 1, a₂ = 4, a₃ = 13, and so on.

Exam Tip: If you are tasked with finding the limit L of a recursive sequence a_n+1 = f(a_n), and you are *already certain* that the sequence converges, you can compute L by solving the equation:

L = f(L)

Warning: You must first prove that the sequence actually converges (e.g., by verifying it is monotone and bounded) before assuming the limit exists and solving this algebraic equation.

5. Tails of Sequences

The "tail" of a sequence represents the long-term behavior of the sequence after discarding a finite number of initial terms.

Definition: Let (a_n) be a sequence. For any fixed natural number m ∈ ℕ, the m-tail of (a_n) is the sequence (a_m+k)_k=1^∞ = (a_m+1, a_m+2, a_m+3, ...).

Theorem: Let (a_n) be a sequence of real numbers, and let m ∈ ℕ. The m-tail (a_m+k)_k=1^∞ converges if and only if the original sequence (a_n) converges. In this case, both sequences converge to the exact same limit.

Important Observation: This theorem formally proves that discarding or altering a finite number of terms at the beginning of a sequence has no effect on its convergence or its limit.

6. Bounded Sequences

Boundedness concerns whether the terms of a sequence are kept within finite bounds.

Definitions:

• (a_n) is bounded above if there exists M ∈ ℝ such that a_n ≤ M for all n ∈ ℕ.
• (a_n) is bounded below if there exists m ∈ ℝ such that m ≤ a_n for all n ∈ ℕ.
• (a_n) is bounded if there exists M ≥ 0 such that |a_n| ≤ M for all n ∈ ℕ.

Theorem: Every convergent sequence of real numbers is bounded.

Proof:

1. Suppose (a_n) converges to L.
2. Set ε = 1. By the definition of a limit, there exists N ∈ ℕ such that for all n ≥ N, |a_n - L| < 1.
3. Using the reverse triangle inequality: |a_n| - |L| ≤ |a_n - L| < 1, which implies |a_n| < |L| + 1 for all n ≥ N.
4. Let M = max(|a₁|, |a₂|, ..., |a_N-1|, |L| + 1).
5. Since |a_n| ≤ M for all indices n < N, and |a_n| < |L| + 1 ≤ M for all n ≥ N, it is clear that |a_n| ≤ M for all n ∈ ℕ.
6. Hence, the sequence is bounded.

Common Mistake: Do not assume the converse is true. A bounded sequence is not guaranteed to converge. For example, the bounded sequence a_n = (-1)ⁿ diverges due to oscillation.

7. Limit Theorems

Limit theorems allow us to compute limits algebraically without returning to the ε-N definition each time.

Theorem (Algebraic Properties of Limits): Let (a_n) and (b_n) be sequences of real numbers that converge to L and M, respectively. Let c ∈ ℝ. Then:

• Constant Multiple Rule: lim_n→∞ (c × a_n) = c × L
• Sum Rule: lim_n→∞ (a_n + b_n) = L + M
• Difference Rule: lim_n→∞ (a_n - b_n) = L - M
• Product Rule: lim_n→∞ (a_n × b_n) = L × M
• Quotient Rule: If b_n ≠ 0 for all n and M ≠ 0, then lim_n→∞ (a_n / b_n) = L / M

Example Proof (Sum Rule)

1. Let ε > 0 be given. We must show |(a_n + b_n) - (L + M)| < ε for sufficiently large n.
2. Observe that |(a_n + b_n) - (L + M)| = |(a_n - L) + (b_n - M)| ≤ |a_n - L| + |b_n - M| by the triangle inequality.
3. Since lim a_n = L, there exists N₁ such that for all n ≥ N₁, |a_n - L| < ε/2.
4. Since lim b_n = M, there exists N₂ such that for all n ≥ N₂, |b_n - M| < ε/2.
5. Let N = max(N₁, N₂). For all n ≥ N, both inequalities hold.
6. Therefore: |(a_n + b_n) - (L + M)| < ε/2 + ε/2 = ε. This completes the proof.

8. Squeeze Theorem

The Squeeze Theorem (or Sandwich Theorem) is a powerful tool to evaluate the limit of a sequence by trapping it between two other sequences with a known, identical limit.

Theorem: Let (a_n), (b_n), and (c_n) be sequences of real numbers such that:
a_n ≤ b_n ≤ c_n   for all n ≥ N₀ (for some N₀ ∈ ℕ).
If lim_n→∞ a_n = lim_n→∞ c_n = L, then (b_n) also converges and lim_n→∞ b_n = L.

Proof:

1. Let ε > 0 be given.
2. Since lim a_n = L, there exists N₁ such that L - ε < a_n < L + ε for all n ≥ N₁.
3. Since lim c_n = L, there exists N₂ such that L - ε < c_n < L + ε for all n ≥ N₂.
4. Let N = max(N₀, N₁, N₂). For all n ≥ N, both limit approximations and the squeeze inequalities hold:
L - ε < a_n ≤ b_n ≤ c_n < L + ε.
5. Thus, L - ε < b_n < L + ε, which is equivalent to |b_n - L| < ε.
6. Hence, lim b_n = L.

Practical Application

Calculate lim_n→∞ (cos(n) / n).

• We know that -1 ≤ cos(n) ≤ 1 for all n.
• Divide the entire inequality by n: -1/n ≤ cos(n)/n ≤ 1/n.
• Since lim_n→∞ (-1/n) = 0 and lim_n→∞ (1/n) = 0, the Squeeze Theorem implies lim_n→∞ (cos(n)/n) = 0.

9. Monotone Sequences

Monotone sequences are sequences whose terms move strictly in one direction.

Definitions:

• (a_n) is monotonically increasing if a_n ≤ a_n+1 for all n ∈ ℕ.
• (a_n) is monotonically decreasing if a_n ≥ a_n+1 for all n ∈ ℕ.
• (a_n) is strictly increasing if a_n < a_n+1 for all n ∈ ℕ.
• (a_n) is strictly decreasing if a_n > a_n+1 for all n ∈ ℕ.

A sequence is monotone if it is either monotonically increasing or monotonically decreasing.

To check monotonicity, analyze either the difference of consecutive terms, or their ratio (if terms are positive):

10. Monotone Convergence Theorem (MCT)

While bounded sequences in general do not always converge, adding the condition of monotonicity guarantees convergence.

Monotone Convergence Theorem: A monotone sequence of real numbers is convergent if and only if it is bounded.

• If (a_n) is monotonically increasing and bounded above, then lim_n→∞ a_n = sup{a_n : n ∈ ℕ}.
• If (a_n) is monotonically decreasing and bounded below, then lim_n→∞ a_n = inf{a_n : n ∈ ℕ}.

Proof Outline (for an increasing sequence):

1. Let S = {a_n : n ∈ ℕ}. Since the sequence is bounded above, S is a non-empty set of real numbers bounded above.
2. By the Completeness Axiom of ℝ (the Supremum Property), S has a least upper bound. Let L = sup(S).
3. We will prove that lim_n→∞ a_n = L. Let ε > 0 be given.
4. Since L = sup(S), L - ε is not an upper bound of S. Thus, there exists some N ∈ ℕ such that a_N > L - ε.
5. Since the sequence is monotonically increasing, for all n ≥ N, we have a_n ≥ a_N > L - ε.
6. Because L is an upper bound of S, we also have a_n ≤ L < L + ε for all n.
7. Therefore, for all n ≥ N, L - ε < a_n < L + ε, which means |a_n - L| < ε.
8. Thus, lim_n→∞ a_n = L.

Example Problem Using MCT

Problem: Let a₁ = 1, and a_n+1 = √(3 + 2a_n). Prove that (a_n) converges and find its limit.

1. Boundedness: Prove by induction that 0 < a_n < 3 for all n.
   - Base case: a₁ = 1, which fits 0 < 1 < 3.
   - Inductive step: Assume 0 < a_k < 3. Then 0 < 2a_k < 6, so 3 < 3 + 2a_k < 9. Taking square roots gives √3 < a_k+1 < 3. Thus, 0 < a_k+1 < 3. The sequence is bounded above by 3.
2. Monotonicity: Prove by induction that a_n < a_n+1.
   - Base case: a₁ = 1, a₂ = √5 ≈ 2.23, so a₁ < a₂.
   - Inductive step: Assume a_k < a_k+1. Then 2a_k < 2a_k+1, which means 3 + 2a_k < 3 + 2a_k+1. Thus, √(3 + 2a_k) < √(3 + 2a_k+1), which means a_k+1 < a_k+2. The sequence is monotonically increasing.
3. Convergence: Since the sequence is monotonically increasing and bounded above, it must converge to some limit L by the Monotone Convergence Theorem.
4. Calculate Limit: Take the limit on both sides of the recurrence relation: L = √(3 + 2L) ⇒ L² = 2L + 3 ⇒ L² - 2L - 3 = 0 ⇒ (L - 3)(L + 1) = 0. Since the terms of the sequence are positive, L must be 3.