sequence-limit

name: sequence-limit description: Prove sequence limits using epsilon-N definition, monotone convergence, and squeeze theorem. Triggers on "sequence converges", "limit of sequence", "prove a_n tends to", "epsilon-N", "monotone convergence", "squeeze theorem", "bounded monotone" tags: [proof, analysis]

Sequence Limit

Prove that a sequence converges to a limit using rigorous techniques.

When to Use

• Establishing that a sequence has a specific limit
• Proving convergence when the limit is unknown
• Applying the squeeze theorem for bounded sequences
• Working with recursively defined sequences

Step 1: Identify the Approach

• Write the sequence explicitly: (a_n)_{n=1}^infty
• Conjecture the limit L (if not given)
• Choose the appropriate technique

Technique Selection

Step 2: The Epsilon-N Definition

For lim_{n->infty} a_n = L:

Formal statement: For all epsilon > 0, exists N in Naturals such that for all n > N, |a_n - L| < epsilon

Proof Template

Let epsilon > 0 be given.
We need to find N such that n > N implies |a_n - L| < epsilon.

[Solve |a_n - L| < epsilon for n]

Choose N = [formula depending on epsilon].

Verification:
For n > N:
  |a_n - L| = ...
            <= ...
            < epsilon

Therefore lim a_n = L.

Finding N

1. Start with |a_n - L| < epsilon
2. Simplify the left side
3. Solve for n
4. N = ceiling of the solution (ensure N is a natural number)

Step 3: Monotone Convergence Theorem

Theorem: Every bounded monotone sequence converges.

• Increasing and bounded above => converges to sup
• Decreasing and bounded below => converges to inf

Proof Strategy

1. Show monotone:

   • Increasing: a_{n+1} >= a_n for all n, OR
   • Decreasing: a_{n+1} <= a_n for all n

2. Show bounded:

   • Find M such that |a_n| <= M for all n, OR
   • Find upper bound (if increasing) or lower bound (if decreasing)

3. Find the limit:

   • If a_n = f(a_{n-1}), let L = lim a_n
   • Then L = f(L) (taking limits of both sides)
   • Solve for L

Step 4: Squeeze Theorem

Theorem: If a_n <= b_n <= c_n for all n >= N_0, and lim a_n = lim c_n = L, then lim b_n = L.

Proof Strategy

1. Find lower bound sequence a_n with known limit L
2. Find upper bound sequence c_n with same limit L
3. Verify a_n <= b_n <= c_n
4. Conclude lim b_n = L

Step 5: Lean Formalization

import Mathlib.Topology.Sequences
import Mathlib.Topology.Order.MonotoneConvergence

-- Epsilon-N definition
theorem seq_limit_def (a : Nat -> Real) (L : Real) :
    Filter.Tendsto a Filter.atTop (nhds L) <->
    forall epsilon > 0, exists N, forall n >= N, |a n - L| < epsilon := by
  sorry -- Use Metric.tendsto_atTop

-- Monotone convergence (increasing, bounded above)
theorem mono_cvg (a : Nat -> Real) (M : Real)
    (h_mono : Monotone a)
    (h_bdd : forall n, a n <= M) :
    exists L, Filter.Tendsto a Filter.atTop (nhds L) := by
  sorry -- Use tendsto_atTop_ciSup

-- Squeeze theorem
theorem squeeze (a b c : Nat -> Real) (L : Real)
    (hab : forall n, a n <= b n)
    (hbc : forall n, b n <= c n)
    (ha : Filter.Tendsto a Filter.atTop (nhds L))
    (hc : Filter.Tendsto c Filter.atTop (nhds L)) :
    Filter.Tendsto b Filter.atTop (nhds L) := by
  exact tendsto_of_tendsto_of_tendsto_of_le_of_le ha hc hab hbc

Key Mathlib lemmas:

• Metric.tendsto_atTop - epsilon-N characterization
• tendsto_atTop_ciSup - monotone convergence
• tendsto_of_tendsto_of_tendsto_of_le_of_le - squeeze theorem

Common Sequence Limits

Worked Examples

Example 1: Prove lim_{n->infty} 1/n = 0

Method: Epsilon-N definition

Let epsilon > 0 be given. We need |1/n - 0| = 1/n < epsilon. This holds when n > 1/epsilon.

Choose N = ceiling(1/epsilon).

For n > N: 1/n < 1/N <= epsilon

Therefore lim 1/n = 0.

Example 2: Prove lim_{n->infty} (1 + 1/n)^n = e

Method: Monotone convergence

Let a_n = (1 + 1/n)^n.

Monotonicity: a_n is increasing (proven via AM-GM or logarithmic analysis).

Boundedness: a_n < 3 for all n. Proof: By binomial theorem, a_n = sum_{k=0}^n C(n,k) (1/n)^k = sum_{k=0}^n (1/k!) * [n(n-1)...(n-k+1)/n^k] < sum_{k=0}^n 1/k! < 1 + 1 + 1/2 + 1/4 + 1/8 + ... = 3

By monotone convergence, a_n converges. Define e = lim a_n.

Example 3: Prove lim_{n->infty} n^{1/n} = 1

Method: Squeeze theorem

For n >= 1, clearly n^{1/n} >= 1.

For upper bound, let n^{1/n} = 1 + h_n where h_n >= 0. Then n = (1 + h_n)^n >= 1 + n*h_n + C(n,2)*h_n^2 >= n(n-1)h_n^2/2

So h_n^2 <= 2/(n-1), giving h_n <= sqrt(2/(n-1)).

Thus: 1 <= n^{1/n} <= 1 + sqrt(2/(n-1))

As n -> infty, the upper bound -> 1. By squeeze theorem, lim n^{1/n} = 1.

Example 4: Recursive sequence a_1 = 1, a_{n+1} = sqrt(2 + a_n)

Method: Monotone convergence + solve for limit

Claim: a_n is increasing and bounded above by 2.

Boundedness: By induction. a_1 = 1 < 2. If a_n < 2, then a_{n+1} = sqrt(2 + a_n) < sqrt(4) = 2.

Monotonicity: a_2 = sqrt(3) > 1 = a_1. If a_{n+1} > a_n, then a_{n+2} = sqrt(2 + a_{n+1}) > sqrt(2 + a_n) = a_{n+1}.

By monotone convergence, let L = lim a_n.

Taking limits: L = sqrt(2 + L) L^2 = 2 + L L^2 - L - 2 = 0 (L - 2)(L + 1) = 0

Since a_n > 0, we have L = 2.

Output Format

**Sequence:** a_n = [formula]

**Claim:** lim_{n->infty} a_n = L

**Method:** [Epsilon-N / Monotone Convergence / Squeeze Theorem]

**Proof:**

[For Epsilon-N:]
Let epsilon > 0. Choose N = [formula].
For n > N: |a_n - L| = ... < epsilon.

[For Monotone Convergence:]
Monotonicity: [show a_{n+1} >= a_n or <=]
Boundedness: [show |a_n| <= M]
Limit equation: [if recursive, solve L = f(L)]

[For Squeeze:]
Lower bound: a_n >= [sequence] -> L
Upper bound: a_n <= [sequence] -> L

**Conclusion:** lim a_n = L. //

Common Pitfalls

1. N must be a natural number: Always take ceiling when N = f(epsilon) is not an integer.

2. Verifying monotonicity: a_{n+1} - a_n >= 0 OR a_{n+1}/a_n >= 1 (for positive terms).

3. Multiple solutions for L: When solving L = f(L), check which solution is consistent with the sequence's sign/bounds.

4. Squeeze requires same limit: Both bounding sequences must converge to the SAME limit.

5. Forgetting base case: In recursive sequences, verify the first few terms satisfy claimed bounds.

6. Index confusion: Be careful with n > N vs n >= N. The definition uses "for all n sufficiently large."