Sequences of Real Numbers ((𝑎_𝑛)_𝑛∊ℕ, 𝑎: ℕ → ℝ)

is a map from natural numbers ℕ to real numbers ℝ (𝑎: ℕ → ℝ)

Sequences of Real Numbers - Syntax

𝑎: ℕ → ℝ
(𝑎₁, 𝑎₂, … )
(𝑎_𝑛)_𝑛∊ℕ

Sequences of Real Numbers - Examples

(𝑎_𝑛)_𝑛∊ℕ	(1)_𝑛∊ℕ	(1, 1, 1, 1, …)
(𝑎_𝑛)_𝑛∊ℕ	((-1)^𝑛)_𝑛∊ℕ	(-1, 1, -1, 1, …)
(𝑎_𝑛)_𝑛∊ℕ	(1/𝑛)_𝑛∊ℕ	(1/1, 1/2, 1/3, 1/4, …)
(𝑎_𝑛)_𝑛∊ℕ	(2^𝑛)_𝑛∊ℕ	(2, 4, 8, 16, 32, …)

Sequences of Real Numbers - Properties

Property	Definition	Examples	Proof
Convergent	a sequence (𝑎_𝑛)_𝑛∊ℕ is convergent if: ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<𝜀	(1)_𝑛∊ℕ	proof Proof by Confirmation Assume the sequence (1)_𝑛∊ℕ is convergent: ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<𝜀 ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1-𝑎̃\|<𝜀 Let 𝑎̃=1: ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1-1\|<𝜀 ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|0\|<𝜀 ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → 0<𝜀 Choose 𝜀 to be a constant 𝑐 greater than 0: ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → 0 < 𝑐 > 0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → true ∃𝛮∊ℕ ∀𝑛∊ℕ : true # conditional simplifies to true true Proof by Contradiction Assume the sequence (1)_𝑛∊ℕ is divergent: ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|𝑎_𝑛-𝑎̃\|≥𝜀 ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1-𝑎̃\|≥𝜀 Choose 𝑎̃ = 1: ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1-1\|≥𝜀 ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|0\|≥𝜀 ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND 0≥𝜀 Let 𝜀 be some constant 𝑐 greater than 0: ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND 0≥𝑐>0 Choose 𝛮 = 1: ∃𝑛∊ℕ : 𝑛≥1 AND 0≥𝑐>0 Let 𝑛 be a number 𝑛≥1 0 ≥ 𝑐 > 0 0 > 0 Thus contradiction
Convergent		(1/𝑛)_𝑛∊ℕ	proof Proof by Confirmation Assume the sequence (1/𝑛)_𝑛∊ℕ is convergent: ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<𝜀 ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1/𝑛-𝑎̃\|<𝜀 Let 𝑎̃=0: ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1/𝑛-0\|<𝜀 ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1/𝑛\|<𝜀 Let 𝜀 to be a constant 𝑐 greater than 0: ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|1/𝑛\| < 𝑐 > 0 Let 𝛮 = 1: ∀𝑛∊ℕ : 𝑛≥1 → \|1/𝑛\| < 𝑐 > 0 Let 𝑛 go to ∞: ∞≥1 → \|1/∞\| < 𝑐 > 0 true → \|0\| < 𝑐 > 0 true → 0 < 𝑐 > 0 true → true true # conditional simplifies to true Proof by Contradiction Assume the sequence (1/𝑛)_𝑛∊ℕ is divergent: ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|𝑎_𝑛-𝑎̃\|≥𝜀 ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1/𝑛-𝑎̃\|≥𝜀 Choose 𝑎̃ = 0: ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1/𝑛-0\|≥𝜀 ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1/𝑛\|≥𝜀 Let 𝜀 be some constant 𝑐 greater than 0: ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|1/𝑛\|≥𝑐>0 Choose 𝛮 = 1: ∃𝑛∊ℕ : 𝑛≥1 AND \|1/𝑛\|≥𝑐>0 Let 𝑛 be a number 𝑛≥1 \|1/𝑛\| ≥ 𝑐 > 0 Let 𝑛 go to infinity: \|1/∞\| ≥ 𝑐 > 0 \|0\| ≥ 𝑐 > 0 0 > 0 Thus contradiction
Divergent	a sequence (𝑎_𝑛)_𝑛∊ℕ is divergent if it is not convergent: ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|𝑎_𝑛-𝑎̃\|≥𝜀	((-1)^𝑛)_𝑛∊ℕ	proof Proof by Confirmation Assume ((-1)^𝑛)_𝑛∊ℕ is divergent: ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|𝑎_𝑛-𝑎̃\|≥𝜀 ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-𝑎̃\|≥𝜀 Let 𝑎̃∊ℝ and 𝜀>0: ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-𝑎̃\|≥𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-𝑎̃\|>0 Let 𝛮∊ℕ: ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-𝑎̃\|>0 If 𝑎̃ = ? 𝑎̃ = 1 ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-1\|>0 Let 𝑛 be an odd number greater than 𝑁: true AND \|-1-1\|>0 \|-2\|>0 true 𝑎̃ = -1 ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛+1\|>0 Let 𝑛 be an even number greater than 𝑁: true AND \|1+1\|>0 \|2\|>0 true 𝑎̃ = any other number ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|(-1)^𝑛-𝑎̃\|>0 Let 𝑛 be an even number greater than 𝑁: true AND \|(-1)^𝑛-𝑎̃\|>0 true AND \|1-𝑎̃\|>0 \|1-𝑎̃\|>0 true # since 𝑎̃ ≠ 1 𝑛≥𝑁: true AND \|(-1)^𝑛-𝑎̃\|>0 \|(-1)^𝑛-𝑎̃\| > 0 Proof by Contradiction Proof #1 ^𝑛)_𝑛∊ℕ is convergent to 𝑎̃∊ℝ. Then: ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<𝜀 Let 𝑎̃=𝑎̃ and 𝜀=1: ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<1 then both: \|𝑎_𝑛 - 𝑎̃\| < 1 \|𝑎_𝑛+1 - 𝑎̃\| < 1 adding the above inequalities leads to: \|𝑎_𝑛 - 𝑎̃\| + \|𝑎_𝑛+1 - 𝑎̃\| < 2 \|1 - 𝑎̃\| + \|(-1) - 𝑎̃\| < 2 \|1 - 𝑎̃\| + \|-(1 + 𝑎̃)\| < 2 \|1 - 𝑎̃\| + \|1 + 𝑎̃\| < 2 # absolute value Let: 2 = \|1 - (-1)\| 2 = \|1 - 𝑎̃ + 𝑎̃ - (-1)\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ - (-1)\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ + 1\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ + 1\| < 2 # above inequality 2 < 2 # contradiction Assume ((-1) Proof #2 ^𝑛)_𝑛∊ℕ is convergent to 𝑎̃∊ℝ. Then: ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|𝑎_𝑛-𝑎̃\|<𝜀 ∃𝑎̃∊ℝ ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛∊ℕ : 𝑛≥𝑁 → \|(-1)^𝑛-𝑎̃\|<𝜀 Let 𝑎̃=𝑎̃, 𝜀=1, and 𝛮∊ℕ: ∀𝑛∊ℕ : 𝑛≥𝑁 → \|(-1)^𝑛-𝑎̃\|<1 Let 𝑛=𝑁 and 𝑛=𝑁+1 then both: 𝑁≥𝑁 → \|(-1)^𝑁-𝑎̃\|<1 𝑁+1≥𝑁 → \|(-1)^𝑁+1-𝑎̃\|<1 Both statements above can be simplified to: true → \|(-1)^𝑁-𝑎̃\|<1 true → \|(-1)^𝑁+1-𝑎̃\|<1 Both statements above can be simplified again to: \|(-1)^𝑁-𝑎̃\|<1 \|(-1)^𝑁+1-𝑎̃\|<1 If 𝑁 is odd then 𝑁+1 is even, and vice versa. Thus yielding the following inequalities: \|-1 - 𝑎̃\| < 1 \|1 - 𝑎̃\| < 1 adding the above inequalities leads to: \|1 - 𝑎̃\| + \|-1 - 𝑎̃\| < 2 \|1 - 𝑎̃\| + \|-(1 + 𝑎̃)\| < 2 \|1 - 𝑎̃\| + \|1 + 𝑎̃\| < 2 Let: 2 = \|1 - (-1)\| 2 = \|1 - 𝑎̃ + 𝑎̃ - (-1)\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ - (-1)\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ + 1\| 2 ≤ \|1 - 𝑎̃\| + \|𝑎̃ + 1\| < 2 # substituting above inequality 2 < 2 # contradiction Assume ((-1)
Divergent		(2^𝑛)_𝑛∊ℕ	proof Proof by Confirmation Assume (2^𝑛)_𝑛∊ℕ is divergent: ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|𝑎_𝑛-𝑎̃\|≥𝜀 ∀𝑎̃∊ℝ ∃𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|2^𝑛-𝑎̃\|≥𝜀 Let 𝑎̃∊ℝ and 𝜀>0: ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|2^𝑛-𝑎̃\|≥𝜀>0 ∀𝛮∊ℕ ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|2^𝑛-𝑎̃\|>0 Let 𝛮∊ℕ: ∃𝑛∊ℕ : 𝑛≥𝑁 AND \|2^𝑛-𝑎̃\|>0 Choose 𝑛 such that 2^𝑛> 𝑎̃ AND 𝑛≥𝑁: true AND \|2^𝑛-𝑎̃\|>0 \|2^𝑛-𝑎̃\|>0 true Proof by Contradiction Assume (2^𝑛)_𝑛∊ℕ is convergent to 𝑎̃∊ℝ. Then: ∀𝜀>0 ∃𝛮∊ℕ ∀𝑛≥𝑁 : \|𝑎_𝑛 - 𝑎̃\| < 𝜀 Let 𝜀=1, so there exists an 𝑁 such that: \|𝑎_𝑁 - 𝑎̃\| < 1 \|𝑎_𝑁_+𝑛 - 𝑎̃\| < 1 \|2^𝑁+𝑛 - 𝑎̃\| < 1 \|2^𝑁+∞ - 𝑎̃\| < 1 # let 𝑛 = ∞ \|2^∞ - 𝑎̃\| < 1 \|∞\| < 1 ∞ < 1 # contradiction
Bounded	a sequence (𝑎_𝑛)_𝑛∊ℕ is bounded if: ∃𝜀∊ℝ ∀𝑛∊ℕ : \|𝑎_𝑛\| ≤ 𝜀	((-1)^𝑛)_𝑛∊ℕ	proof Proof by Confirmation Assume ((-1)^𝑛)_𝑛∊ℕ is bounded: ∃𝜀∊ℝ ∀𝑛∊ℕ : \|(-1)^𝑛\| ≤ 𝜀 Let 𝜀=2, thus: ∀𝑛∊ℕ : \|(-1)^𝑛\| ≤ 2 Proof by Contradiction Assume ((-1)^𝑛)_𝑛∊ℕ is unbounded: ∀𝜀∊ℝ ∃𝑛∊ℕ : \|𝑎_𝑛\| > 𝜀 ∀𝜀∊ℝ ∃𝑛∊ℕ : \|(-1)^𝑛\| > 𝜀 Let 𝜀=2: ∃𝑛∊ℕ : \|(-1)^𝑛\| > 2 Thus we have both: \|-1\| > 2 \|1\| > 2 These are both contradictions
Unbounded	a sequence (𝑎_𝑛)_𝑛∊ℕ is unbounded if it is not bounded: ∀𝜀∊ℝ ∃𝑛∊ℕ : \|𝑎_𝑛\| > 𝜀	(2^𝑛)_𝑛∊ℕ	proof Proof by Confirmation Assume is (2^𝑛)_𝑛∊ℕ unbounded: ∀𝜀∊ℝ ∃𝑛∊ℕ : \|2^𝑛\| > 𝜀 Let 𝜀 be a constant 𝑐: ∃𝑛∊ℕ : \|2^𝑛\| > 𝑐 ∃𝑛∊ℕ : 2^𝑛 > 𝑐 ∃𝑛∊ℕ : 𝑙𝑔(2^𝑛) > 𝑙𝑔(𝑐) ∃𝑛∊ℕ : 𝑛·𝑙𝑔(2) > 𝑙𝑔(𝑐) ∃𝑛∊ℕ : 𝑛 > 𝑙𝑔(𝑐) Proof by Contradiction Assume (2^𝑛)_𝑛∊ℕis bounded: ∃𝜀∊ℝ ∀𝑛∊ℕ : \|𝑎_𝑛\| ≤ 𝜀 ∃𝜀∊ℝ ∀𝑛∊ℕ : \|2^𝑛\| ≤ 𝜀 ∃𝜀∊ℝ ∀𝑛∊ℕ : \|2^∞\| ≤ 𝜀 # let 𝑛 = ∞ ∃𝜀∊ℝ ∀𝑛∊ℕ : ∞ ≤ 𝜀 𝜀 is greater than or equal to ∞, but ∞ does not exist in ℝ. Thus contradiction

theorems:

if a sequence is convergent then it is also bounded
if a sequence is unbounded then it is also divergent

TODO

Theorem on Limits

Given two convergent sequences (𝑎_𝑛)_𝑛∊ℕ and (𝑏_𝑛)_𝑛∊ℕ then:

𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛 + 𝑏_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) + 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛)
𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛 · 𝑏_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) · 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛)
𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛 / 𝑏_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) / 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛) # 𝑏_𝑛≠ 0

Monotonicity

Given two convergent sequences (𝑎_𝑛)_𝑛∊ℕ and (𝑏_𝑛)_𝑛∊ℕ:

if 𝑎_𝑛≤ 𝑏_𝑛 for all 𝑛∊ℕ → 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) ≤ 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛)

Monotonically Increasing/Decreasing

A sequence (𝑎_𝑛)_𝑛∊ℕ is:

monotonically increasing if: 𝑎_𝑛≤ 𝑎_𝑛+1 for all 𝑛
monotonically decreasing if: 𝑎_𝑛≥ 𝑎_𝑛+1 for all 𝑛

Strictly Monotonically Increasing/Decreasing

A sequence (𝑎_𝑛)_𝑛∊ℕ is:

strictly monotonically increasing if: 𝑎_𝑛< 𝑎_𝑛+1 for all 𝑛
strictly monotonically decreasing if: 𝑎_𝑛> 𝑎_𝑛+1 for all 𝑛

Bounded From Above/Below - Bounded

A sequence (𝑎_𝑛)_𝑛∊ℕ is:

bounded from above if the set {𝑎_𝑛}_𝑛∊ℕ has an upper bound
bounded from below if the set {𝑎_𝑛}_𝑛∊ℕ has a lower bound
bounded if the set {𝑎_𝑛}_𝑛∊ℕ has an upper bound and lower bound

Sandwich Theorem

Given two convergent sequences (𝑎_𝑛)_𝑛∊ℕ and (𝑏_𝑛)_𝑛∊ℕ, and a sequence (𝑐_𝑛)_𝑛∊ℕ:

if 𝑎_𝑛≤ 𝑏_𝑛 ≤ 𝑐_𝑛for all 𝑛∊ℕ AND 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛) → (𝑐_𝑛)_𝑛∊ℕ is convergent with 𝑙𝑖𝑚_𝑛→∞(𝑐_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑎_𝑛) = 𝑙𝑖𝑚_𝑛→∞(𝑏_𝑛)

Cauchy Sequence

A sequence (𝑎_𝑛)_𝑛∊ℕ is a Cauchy sequence if:

∀𝜀>0 ∃𝑁∊ℕ ∀𝑛,𝑚≥𝑁 : |𝑎_𝑛 - 𝑎_𝑚| < 𝜀

Info

For sequences of real numbers:

Cauchy sequence ↔ convergent sequence

Dedekind Completeness

If subset 𝑆⊆ℝ is bounded from above, then 𝑠𝑢𝑝𝑟𝑒𝑚𝑢𝑚(𝑆) exists
If subset 𝑆⊆ℝ is bounded from below, then 𝑖𝑛𝑓𝑖𝑚𝑢𝑚(𝑆) exists

Monotone Convergence Criterion

A sequence (𝑎_𝑛)_𝑛∊ℕ is convergent if both:

monotonically decreasing (i.e. 𝑎_𝑛≥𝑎_𝑛+1 for all 𝑛)
bounded from below (i.e. the set {𝑎_𝑛}_𝑛∊ℕ has a lower bound)

A sequence (𝑎_𝑛)_𝑛∊ℕ is convergent if both:

monotonically increasing (i.e. 𝑎_𝑛≤𝑎_𝑛+1 for all 𝑛)
bounded from above (i.e. the set {𝑎_𝑛}_𝑛∊ℕ has an upper bound)

Accumulation/Cluster/Partial-Limit Values/Point

𝑎̃∊ℝ is called an accumulation value of a sequence (𝑎_𝑛)_𝑛∊ℕ if either:

there exists a subsequence (𝑎_𝑛𝑘)_𝑘∊ℕ with 𝑙𝑖𝑚_𝑘→∞𝑎_𝑛𝑘 = 𝑎̃ (i.e. subsequence is convergent)
∀𝜀>0: the 𝜀-neighborhood of 𝑎̃ contains infinitely many sequence members of (𝑎_𝑛)_𝑛∊ℕ

Improper Accumulation Value

a sequence (𝑎_𝑛)_𝑛∊ℕ has the improper accumulation value ∞ ↔ (𝑎_𝑛)_𝑛∊ℕ is not bounded from above
a sequence (𝑎_𝑛)_𝑛∊ℕ has the improper accumulation value ∞ ↔ (𝑎_𝑛)_𝑛∊ℕ is not bounded from below

Limit Superior - Limit Inferior

Given a sequence (𝑎_𝑛)_𝑛∊ℕ, an element 𝑎̃ ∊ ℝ∪{∞,-∞} = [-∞,∞] is called:

limit superior of (𝑎_𝑛)_𝑛∊ℕif 𝑎̃ is the largest (improper) accumulation value of (𝑎_𝑛)_𝑛∊ℕ (denoted as 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑎_𝑛)
limit inferior of (𝑎_𝑛)_𝑛∊ℕif 𝑎̃ is the smallest (improper) accumulation value of (𝑎_𝑛)_𝑛∊ℕ(denoted as 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑎_𝑛)

Limit superior relates to supremum:

𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑎_𝑛 = 𝑙𝑖𝑚_𝑛→∞𝑠𝑢𝑝𝑟𝑒𝑚𝑢𝑚{𝑎_𝑘 | 𝑘 ≥ 𝑛}

Limit inferior relates to infimum:

𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑎_𝑛 = 𝑙𝑖𝑚_𝑛→∞𝑖𝑛𝑓𝑖𝑚𝑢𝑚{𝑎_𝑘 | 𝑘 ≥ 𝑛}

Limit Superior/Inferior Algebra:

𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛+𝑏_𝑛) ≤ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛) + 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑏_𝑛)
𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛·𝑏_𝑛) ≤ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛) · 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑏_𝑛) # if 𝑎_𝑛,𝑏_𝑛≥ 0
𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛+𝑏_𝑛) ≥ 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛) + 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑏_𝑛)
𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛·𝑏_𝑛) ≥ 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛) · 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑏_𝑛) # if 𝑎_𝑛,𝑏_𝑛≥ 0

Limit Superior/Inferior Algebra Examples:

Example #1

Given the following sequences:

(𝑎_𝑛)_𝑛∊ℕ = (1, -1, 1, -1, 1, …)

(𝑏_𝑛)_𝑛∊ℕ = (0, 2, 0, 2, 0, …)

(𝑎_𝑛+ 𝑏_𝑛)_𝑛∊ℕ = (1, 1, 1, 1, 1, …)

(𝑎_𝑛· 𝑏_𝑛)_𝑛∊ℕ = (0, -2, 0, -2, 0, …)

Then:

1 = 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛+𝑏_𝑛) ≤ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛) + 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑏_𝑛) = 1 + 2 = 3

1 = 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛+𝑏_𝑛) ≥ 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛) + 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑏_𝑛) = -1 + 0 = -1

Example #2

Given the following sequences:

(𝑎_𝑛)_𝑛∊ℕ = (1, 0, 1, 0, 1, …)

(𝑏_𝑛)_𝑛∊ℕ = (0, 2, 0, 2, 0, …)

(𝑎_𝑛· 𝑏_𝑛)_𝑛∊ℕ = (0, 0, 0, 0, 0, …)

Then:

0 = 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛·𝑏_𝑛) ≤ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑎_𝑛) · 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞(𝑏_𝑛) = 1 · 2 = 2

0 = 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛·𝑏_𝑛) ≥ 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑎_𝑛) · 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞(𝑏_𝑛) = 0 · 0 = 0

Convergent/Divergent vs Limit Superior/Inferior

(𝑎_𝑛)_𝑛∊ℕ is convergent ↔ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑎_𝑛= 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑎_𝑛 ∉ {±∞}
(𝑎_𝑛)_𝑛∊ℕ is divergent to ∞ ↔ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑎_𝑛= 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑎_𝑛 = ∞
(𝑎_𝑛)_𝑛∊ℕ is divergent to -∞ ↔ 𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑎_𝑛= 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑎_𝑛 = -∞

Bolzano-Weierstrass Theorem

Bolzano-Weierstrass Theorem:

sequence (𝑎_𝑛)_𝑛∊ℕ is bounded → (𝑎_𝑛)_𝑛∊ℕ has at least 1 accumulation value

Proof:

Click here to expand...

TODO

Divergent to Infinity

divergent to infinity ↔ 𝑙𝑖𝑚_𝑛→∞𝑎_𝑛 = ∞ ↔ ∀𝑐>0 ∃𝑁∊ℕ ∀𝑛≥𝑁 : 𝑎_𝑛>𝑐
divergent to -infinity ↔ 𝑙𝑖𝑚_𝑛→∞𝑎_𝑛 = -∞ ↔ ∀𝑐<0 ∃𝑁∊ℕ ∀𝑛≥𝑁 : 𝑎_𝑛<𝑐

Epsilon/𝜀-Neighborhoods

For 𝜀>0, (𝑥-𝜀,𝑥+𝜀) = 𝐵_𝜀(𝑥) is the epsilon-neighborhood of 𝑥

𝑆⊆ℝ is called a neighborhood of 𝑥 if:

∃𝜀>0 : 𝐵_𝜀(𝑥)⊆𝑆

Open Sets - Closed Sets

𝑆⊆ℝ is called open (in ℝ) if:

∀𝑥∊𝑆 ∃𝜀>0 : 𝐵_𝜀(𝑥)⊆𝑆

𝑆⊆ℝ is called closed (in ℝ) if:

𝑆^c = ℝ\𝑆 is open

Open and Closed Sets Examples:

⦰, ℝ are both open and closed
[-2, 2] is closed but not open
(-2, 2) is open but not closed
(-2,2] is neither open nor closed

Closed Sets vs Convergent Sequences

both statements are equivalent:

𝑆⊆ℝ is closed ↔ for all convergent sequences (𝑎_𝑛)_𝑛∊ℕ with 𝑎_𝑛∊𝑆 for all 𝑛∊ℕ we have: 𝑙𝑖𝑚_𝑛→∞𝑎_𝑛 ∊ 𝑆
𝑆⊆ℝ is closed ↔ any convergent sequences (𝑎_𝑛)_𝑛∊ℕ⊆𝑆 has limit point 𝑎̃∊𝑆

Compact Sets vs Convergent Subsequences

both statements are equivalent:

𝑆⊆ℝ is compact ↔ for all sequences (𝑎_𝑛)_𝑛∊ℕ with 𝑎_𝑛∊𝑆 for all 𝑛∊ℕ, there exists a convergent subsequence (𝑎_𝑛𝑘)_𝑘∊ℕ with 𝑙𝑖𝑚_𝑘→∞𝑎_𝑛𝑘 ∊ 𝑆
𝑆⊆ℝ is compact ↔ any sequence (𝑎_𝑛)_𝑛∊ℕ⊆𝑆 has an accumulation value 𝑎̃∊𝑆

Compact Sets Examples:

⦰ is compact
{5} is compact
ℝ is not compact (is closed though) # for example (𝑎_𝑛)_𝑛∊ℕ = (𝑛)_𝑛∊ℕ has no accumulation value 𝑎̃∊𝑆
[𝑐,𝑑] is compact
proof
Let (𝑎_𝑛)_𝑛∊ℕ⊆[𝑐,𝑑] thus:
- (𝑎_𝑛)_𝑛∊ℕ⊆[𝑐,𝑑] → (𝑎_𝑛)_𝑛∊ℕ is a bounded sequence
- (𝑎_𝑛)_𝑛∊ℕ⊆[𝑐,𝑑] → (𝑎_𝑛)_𝑛∊ℕ is has an accumulation value 𝑎̃∊ℝ # via Bolzano-Weierstrass Theorem
- (𝑎_𝑛)_𝑛∊ℕ⊆[𝑐,𝑑] → accumulation value actually satisfies 𝑎̃∊[𝑐,𝑑] # [𝑐,𝑑] is closed

Heine-Borel Theorem

For 𝑆⊆ℝ, we have:

𝑆 is compact ↔ 𝑆 is bounded and closed

Proof:

𝑆 is bounded and closed → 𝑆 is compact

Assume 𝑆⊆ℝ is bounded and closed.

Let (𝑎_𝑛)_𝑛∊ℕ be a sequence in 𝑆 thus:

(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 → (𝑎_𝑛)_𝑛∊ℕ⊆𝑆

(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 → (𝑎_𝑛)_𝑛∊ℕ is a bounded sequence

(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 → (𝑎_𝑛)_𝑛∊ℕ has an accumulation value 𝑎̃∊ℝ # via Bolzano-Weierstrass Theorem

(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 → accumulation value actually satisfies 𝑎̃∊𝑆 # because 𝑆 is closed

𝑆 is compact → 𝑆 is closed

Assume 𝑆⊆ℝ is compact. Prove that for any convergent sequences (𝑎_𝑛)_𝑛∊ℕ with 𝑎_𝑛∊𝑆 for all 𝑛∊ℕ we have: 𝑙𝑖𝑚_𝑛→∞𝑎_𝑛∊𝑆

Let:

(𝑎_𝑛)_𝑛∊ℕ be a convergent sequence in 𝑆

(𝑎_𝑛)_𝑛∊ℕ has an accumulation value 𝑎̃∊𝑆 # via 𝑆 is compact

𝑙𝑖𝑚_𝑛→∞𝑎_𝑛 = 𝑎̃ ∊ 𝑆 # Every convergent sequence has only 1 accumulation value

𝑙𝑖𝑚_𝑛→∞𝑎_𝑛 ∊ 𝑆

𝑆 is compact → 𝑆 is bounded

Proof by Contradiction

𝑆 is not bounded → 𝑆 is not compact

Bounded Definition

a set is bounded if it has an upper bound and lower bound

Not Bounded Definition

a set is unbounded if (does not have an upper bound) or (does not have a lower bound)

Compact Definition

∀(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 ∃(𝑎_𝑛𝑘)_𝑘∊ℕ : (𝑎_𝑛𝑘)_𝑘∊ℕis convergent AND 𝑙𝑖𝑚_𝑘→∞𝑎_𝑛𝑘∊𝑆

Not Compact Definition

∃(𝑎_𝑛)_𝑛∊ℕ⊆𝑆 ∀(𝑎_𝑛𝑘)_𝑘∊ℕ : (𝑎_𝑛𝑘)_𝑘∊ℕis not convergent OR 𝑙𝑖𝑚_𝑘→∞𝑎_𝑛𝑘∉𝑆

Assume 𝑆⊆ℝ is unbounded. Prove that 𝑆 is not convergent (i.e. there exists a sequence in 𝑆 such that all subsequences are not convergent)

𝑆⊆ℝ is unbounded → 𝑆⊆ℝ is unbounded

𝑆⊆ℝ is unbounded → There is a sequence (𝑎_𝑛)_𝑛∊ℕ⊆𝑆 such that |𝑎_𝑛|>𝑛 for all 𝑛∊ℕ # 𝑆 does not have a lower bound or upper bound

𝑆⊆ℝ is unbounded → all subsequences of (𝑎_𝑛)_𝑛∊ℕ⊆𝑆 is not convergent

Next Topics

Absolutely Convergent - Absolute Convergence

A series 𝛴_{1≤𝑖≤∞}𝑎_𝑖 absolutely convergent if 𝛴_{1≤𝑖≤∞}|𝑎_𝑖| is convergent

Theorem:

absolute convergence → convergence

Proof:

Click here to expand...

𝛴_{1≤𝑖≤∞}|𝑎_𝑖| is convergent → ∀𝜀>0 ∃𝑁∊ℕ ∀𝑛≥𝑚≥𝑁 : |𝛴_{1≤𝑖≤∞}|𝑎_𝑖|| < 𝜀 # via Cauchy criterion

𝛴_{1≤𝑖≤∞}|𝑎_𝑖| is convergent → ∀𝜀>0 ∃𝑁∊ℕ ∀𝑛≥𝑚≥𝑁 : 𝛴_{1≤𝑖≤∞}|𝑎_𝑖| < 𝜀 # all summands are positive

𝛴_{1≤𝑖≤∞}|𝑎_𝑖| is convergent → ∀𝜀>0 ∃𝑁∊ℕ ∀𝑛≥𝑚≥𝑁 : |𝛴_{1≤𝑖≤∞}𝑎_𝑖| < 𝜀 # triangle inequality |𝛴_{1≤𝑖≤∞}𝑎_𝑖| ≤ 𝛴_{1≤𝑖≤∞}|𝑎_𝑖|

𝛴_{1≤𝑖≤∞}|𝑎_𝑖| is convergent → 𝛴_{1≤𝑖≤∞}𝑎_𝑖 # via Cauchy criterion

Majorant Criterion

Theorem

Let 𝛴_{1≤𝑖≤∞}𝑎_𝑖be a series. If there is 𝑛₀∊ℕ and a convergent series 𝛴_{1≤𝑖≤∞}𝑏_𝑖satisfying the following for all 𝑖≥𝑛₀:

𝑏_𝑖≥0
|𝑎_𝑖| ≤ 𝑏_𝑖

Then 𝛴_{1≤𝑖≤∞}𝑎_𝑖 is an absolutely convergent series

Proof

Click here to expand...

Apply the Cauchy criterion to series 𝛴_{1≤𝑖≤∞}𝑏_𝑖:

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → ∀𝜀>0 ∃𝑁≥ℕ ∀𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}𝑏_𝑖| < 𝜀 # via the Cauchy criterion

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → ∀𝜀>0 ∃𝑁≥𝑛₀ ∀𝑛≥𝑚≥𝑁 : |𝛴_{𝑚≤𝑖≤𝑛}𝑏_𝑖| < 𝜀 # choose an 𝑁≥𝑛₀

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → ∀𝜀>0 ∃𝑁≥𝑛₀ ∀𝑛≥𝑚≥𝑁 : 𝛴_{𝑚≤𝑖≤𝑛}𝑏_𝑖 < 𝜀 # because 𝑏_𝑖≥0

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → ∀𝜀>0 ∃𝑁≥𝑛₀ ∀𝑛≥𝑚≥𝑁 : 𝛴_{𝑚≤𝑖≤𝑛}|𝑎_𝑖| < 𝜀 # because |𝑎_𝑖| ≤ 𝑏_𝑖

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → 𝛴_{𝑚≤𝑖≤𝑛}|𝑎_𝑖| is a convergent series # via the Cauchy criterion

𝛴_{1≤𝑖≤∞}𝑏_𝑖 is a convergent series → 𝛴_{𝑚≤𝑖≤𝑛}𝑎_𝑖 is an absolutely convergent series # by definition of absolutely convergent series

Minorant Criterion

Theorem

Let 𝛴_{1≤𝑖≤∞}𝑎_𝑖be a series with 𝑎_𝑖≥0. If there is 𝑛₀∊ℕ and a divergent series 𝛴_{1≤𝑖≤∞}𝑏_𝑖satisfying the following for all 𝑖≥𝑛₀:

𝑏_𝑖≥ 0
𝑎_𝑖 ≥ 𝑏_𝑖

Then 𝛴_{1≤𝑖≤∞}𝑎_𝑖 is a divergent series

Proof

Click here to expand...

TODO

Example

Click here to expand...

𝛴_{1≤𝑖≤∞}(1/√𝑖) is divergent because:

√𝑖 ≤ 𝑖 ↔ (1/√𝑖) ≥ (1/𝑖) for all 𝑖

𝛴_{1≤𝑖≤∞}(1/𝑖) is the harmonic series which is known to be divergent

Thus the minorant criterion is satisfied → 𝛴_{1≤𝑖≤∞}(1/√𝑖) is divergent

／var／logmarcus chiu

Explorer

Sequences of Real Numbers (Convergent／Divergent／Bounded／Unbounded Sequences)

Sequences of Real Numbers ((𝑎𝑛)𝑛∊ℕ, 𝑎: ℕ → ℝ)

Sequences of Real Numbers - Syntax

𝑎: ℕ → ℝ

Sequences of Real Numbers - Examples

Sequences of Real Numbers - Properties

Proof by Confirmation

Proof by Contradiction

Proof by Confirmation

Proof by Contradiction

Proof by Confirmation

If 𝑎̃ = ?

Proof by Contradiction

Proof by Confirmation

Proof by Contradiction

Proof by Confirmation

Proof by Contradiction

Proof by Confirmation

Proof by Contradiction

TODO

Theorem on Limits

Monotonicity

Monotonically Increasing/Decreasing

Strictly Monotonically Increasing/Decreasing

Bounded From Above/Below - Bounded

Sandwich Theorem

Cauchy Sequence

Dedekind Completeness

Monotone Convergence Criterion

Accumulation/Cluster/Partial-Limit Values/Point

Improper Accumulation Value

Limit Superior - Limit Inferior

Convergent/Divergent vs Limit Superior/Inferior

Bolzano-Weierstrass Theorem

Divergent to Infinity

Epsilon/𝜀-Neighborhoods

Open Sets - Closed Sets

Closed Sets vs Convergent Sequences

Compact Sets vs Convergent Subsequences

[𝑐,𝑑] is compact

Heine-Borel Theorem

Proof by Contradiction

Next Topics

Absolutely Convergent - Absolute Convergence

Majorant Criterion

Theorem

Proof

Minorant Criterion

Theorem

Proof

Example

Ratio Test (TODO)

Root Test

／var／log marcus chiu

Sequences of Real Numbers ((𝑎_𝑛)_𝑛∊ℕ, 𝑎: ℕ → ℝ)