Cauchy Criterion For Convergence

Cauchy Criterion For Convergence - Proof

Let 𝑠_𝑛 = 𝛴_{1≤𝑖≤𝑛}𝑎_𝑖:

• (𝑠_𝑛)_𝑛∊ℕ is a convergent sequence ↔_completeness (𝑠_𝑛)_𝑛∊ℕ is a Cauchy sequence # via completeness of the real numbers
• (𝑠_𝑛)_𝑛∊ℕ is a convergent sequence ↔ ∀𝜀>0 ∃𝑁∊ℕ ∀𝑛̃,𝑚̃≥𝛮 : |𝑠_𝑛̃ - 𝑠_𝑚̃| < 𝜀 # definition of a Cauchy sequence
• (𝑠_𝑛)_𝑛∊ℕ is a convergent sequence ↔ ∀𝜀>0 ∃𝑁∊ℕ ∀𝑛≥𝑚≥𝛮 : |𝑠_𝑛 - 𝑠_𝑚-1| < 𝜀

Cauchy Criterion For Convergence - Example

Prove (𝛴_{1≤𝑖≤∞}(-1)^𝑖) is not convergent via the Cauchy criterion.

• !(∀𝜀>0 ∃𝑁∊ℕ ∀𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| < 𝜀) → !((𝛴_{1≤𝑖≤∞}(-1)^𝑖) is a convergent series)
• (∃𝜀>0 ∀𝑁∊ℕ ∃𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| ≥ 𝜀) → (𝛴_{1≤𝑖≤∞}(-1)^𝑖) is NOT a convergent series)

Prove the following:

• ∃𝜀>0 ∀𝑁∊ℕ ∃𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| ≥ 𝜀

Choose 𝜀=0.5:

• ∀𝑁∊ℕ ∃𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| ≥ 0.5

Let 𝑁∊ℕ:

• ∃𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| ≥ 0.5

Let 𝑚=𝑁 and 𝑛=𝑁+2:

• |𝛴_{𝑁≤𝑖≤𝑁+2}(-1)^𝑖| ≥ 0.5

If:

• 𝑁 is even → |1 + (-1) + 1| = 1 ≥ 0.5
• 𝑁 is odd → |-1 + 1 + (-1)| = 1 ≥ 0.5

Thus we proved (∃𝜀>0 ∀𝑁∊ℕ ∃𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}(-1)^𝑖| ≥ 𝜀) which means (𝛴_{1≤𝑖≤∞}(-1)^𝑖) is NOT a convergent series)