Prove that if the sequence \(a_n\) converges to \(L\), then \(|a_n|\) converges to \(|L|\).
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε variable {a b : ℕ → ℝ} variable {L : ℝ} example (ha : SeqLim a L) (hb : ∀ n, b n = |a n|) : SeqLim b |L| := by sorry
Prove that if a sequence converges to a nonzero limit, then its terms are eventually bounded below by half the absolute value of the limit; that is, if a sequence \(a_n\) converges to \(L\) with \(L ≠ 0\), then there exists \(k\) such that for all \(n ≥ k\) we have \(|aₙ| ≥ \dfrac{|L|}{2}\).
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε variable {a : ℕ → ℝ} variable {L : ℝ} example (ha : SeqLim a L) (hL : L ≠ 0) : ∃ k, ∀ n ≥ k, |a n| ≥ |L| / 2 := by sorry
This week's challenge consists of proving in Lean 4 that if a sequence \(aₙ\) converges to both \(L\) and \(M\), then \(L = M\).
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε variable {a : ℕ → ℝ} variable {L M : ℝ} example (hL : SeqLim a L) (hM : SeqLim a M) : L = M := by sorry
Prove that a sequence with infinitely many terms whose absolute value is greater than 10 cannot converge to a limit whose absolute value is less than 5.
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε variable {a : ℕ → ℝ} variable {L : ℝ} example (ha : ∀ k, ∃ n ≥ k, |a n| > 10) : ¬ ∃ L, SeqLim a L ∧ |L| < 5 := by sorry
Prove the squeeze theorem; that is, if \(a_n\) and \(c_n\) converge to \(L\) and \(a_n ≤ b_n \leq c_n\) for all \(n\), then \(b_n\) converges to \(L\).
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {a b c : ℕ → ℝ} variable {L : ℝ} def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ N : ℕ, ∀ n ≥ N, |a n - L| < ε example (ha : SeqLim a L) (hc : SeqLim c L) (hab : ∀ n, a n ≤ b n) (hbc : ∀ n, b n ≤ c n) : SeqLim b L := by sorry
Prove that the sequence defined by 1, -1, 1, -1, ... does not converge.
To do so, the following proof in Lean 4 is to be completed:
import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {a : ℕ → ℝ} def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ N : ℕ, ∀ n ≥ N, |a n - L| < ε def SeqConv (a : ℕ → ℝ) : Prop := ∃ L, SeqLim a L example (ha : ∀ n, a n = (-1) ^ n) : ¬ SeqConv a := by sorry
In Lean, a sequence \(a₀, a₁, a₂,...\) can be represented by a function \(a : ℕ → ℝ\) so that \(a(n)\) is \(aₙ\).
The statement that \(L\) is the limit of the sequence \(a\) is defined by
def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε
Prove that if \(aₙ = 1/n\) for all \(n\), then the sequence \(a\) converges to 0.
To do so, complete the following Lean 4 theory:
import Mathlib.Data.Real.Basic import Mathlib.Tactic variable (a : ℕ → ℝ) def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop := ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε example (ha : ∀ n, a n = 1 / n) : SeqLim a 0 := by sorry