Lecture 8: The Squeeze Theorem and Operations Involving Convergent Sequences

18.100A: Complete Lecture Notes Lecture 8: The Squeeze Theorem and Operations Involving Convergent Sequences Facts About Limits Theorem 1 (Squeeze Theorem) Let {an }, {bn }, and {xn } be sequences such that ∀n ∈ N, an ≤ xn ≤ bk . Suppose that {an } and {bn } converge and lim an = x = lim bn . n→∞ n→∞ Therefore, {x} converges and limn→∞ xn = x. Remark 2. We sometimes abbreviate the Squeeze Theorem to ST. Proof : Let  > 0. Since limn→∞ an = x, there exists an M0 ∈ N such that for all n ≥ M0 , |an − x| <  =⇒ x −  < an . Since limn→∞ bn = x, ∃M1 ∈ N such that ∀n ≥ M1 , |bn − x| <  =⇒ bn < x + . Choose M = max{M0 , M1 }. Then, if n ≥ M , then x −  < an ≤ xn ≤ bn < x −  =⇒ |xn − x| < . Therefore, {xn } is convergent and limn→∞ xn = x. Theorem 3 Another way to check that a sequence xn → x, is stated below: lim xn = x ⇐⇒ lim |xn − x| = 0. n→∞ n→∞ Hence, we can consider a sequence like the following: Example 4 Show that n2 = 1. n→∞ n2 + n + 1 lim 1 Proof : We have n2 −n − 1 n+1 n+1 1 = 2 −1 = 2 ≤ 2 = . 2 n +n+1 n +n+1 n +n+1 n +n n Thus, 0≤ n2 n2 1 n2 − 1 ≤ → 0 =⇒ lim −1 =0 2 n→∞ n + n + 1 +n+1 n  by the Squeeze Theorem. Question 5. How do limits interact with ordering? Theorem 6 Let {xn } and {yn } be sequences of real numbers. Then, 1. if {xn } and {yn } are convergent sequences and ∀n ∈ N xn ≤ yn , then limn→∞ xn ≤ limn→∞ yn . 2. if {xn } is a convergent sequence and ∀n ∈ N x ≤ xn ≤ bn then a ≤ limn→∞ xn ≤ b. Proof : 1. Let x = limn→∞ xn and y = limn→∞ yn . Suppose for the sake of contradiction that y < x. Then, ∃M0 ∈ N such that ∀n ≥ M0 x−y |yn − y| < 2 And ∃M1 ∈ N such that for all n ≥ M1 , x−y . |xn − x| < 2 Then, if M = M0 + M1 ≥ max{M0 , M1 }, yM < x−y x+y x−y +y = =x− + x < xM . 2 2 2 However, this would imply that yM < xM which contradicts ∀n ∈ Nxn ≤ yn . 2. Apply part 1 to proof part 2, by considering yn = a ≤ xn ≤ b = zn for all n ∈ N. Question 7. How do limits interact with algebraic operations? Theorem 8 Suppose limn→∞ xn = x and limn→∞ yn = y. Then, 1. {xy + yn }n is convergent and limn→∞ (xn + yn ) = x + y. 2. ∀c ∈ R, {cxn }n is convergent and limn→∞ cxn = cx. 3. {xn · yn } is convergent and limn→∞ xn yn = xy. 4. If ∀n ∈ N, yn 6= 0 and y 6= 0, then {xn /yn }n is convergent and lim n→∞ Proof : 2 xn x = . yn y 1. Let  > 0. Then, since xn → x, ∃M0 ∈ N such that ∀n ≥ M0 , |xn − x| < 2 . Since yn → y, ∃M1 ∈ N such that ∀n ≥ M1 , |yn − y| < 2 . Hence, letting M = max{M0 , M1 }, we get for all n ≥ M , |xn + yn − (x + y)| ≤ |xn − x| + |yn − y| < 2. Let  > 0. Since xn → x, ∃M0 ∈ N such that ∀n ≥ M0 , |xn − x| < |cxn − cx| = |c||xn − x| ≤ since |c| |c|+1   + = . 2 2  |c|+1 . Let M = M0 . Then, ∀n ≥ M , |c| ·< |c| + 1 < 1. 3. Since yn → y, {yn } is bounded. In other words, ∃B ≥ 0 such that ∀n ∈ N, |yn | ≤ B. Then, |xn yn − xy| = |(xn − x)yn + (yn − y)x| ≤ |xn − x||yn + |x||yn − y| ≤ B|xn − x| + |x||yn − y|. Therefore, 0 ≤ |xn yn − xy| ≤ B|xn − x| + |x||yn − y|. Since B|xn − x| + |x||yn − y| → 0, by the Squeeze Theorem limn→∞ |xn yn − xy| = 0. 4. We prove y1n → y1 . We first prove ∃b > 0 such that ∀n ∈ N, |yn | ≥ b. Since yn → y and y 6= 0, ∃M0 ∈ N such that ∀n ≥ M0 , |y| |yn − y| < . 2 By the Triangle Inequality, ∀n ≥ M0 , |y| ≤ |yn − y| + |yn | ≤ |y| |y| + |yn | =⇒ |yn | ≥ . 2 2 n o Let b = min |y1 |, . . . , |yM0 −1 |, |y| 2 . Then, ∀n ∈ N, |yn | ≥ b. Therefore, 0≤ |yn − y| 1 1 1 = ≤ |yn − y|. − yn y |yn ||y| b|y| − 1 = 0. Therefore, limn→∞   y before this (3.), it follows that limn→∞ xn · y1n = xy . By the Squeeze Theorem, limn→∞ 1 yn 1 yn = 1 y. Furthermore, by the proof Remark 9. By induction, one can prove that lim (xn )k = xk . n→∞ Theorem 10 √ If {xn } is a convergent sequence such that ∀n ∈ N, xn ≥ 0, then { xn } is convergent and lim n→∞ q √ xn = lim xn . n→∞ Proof : Let x = limn→∞ xn . 3 Case 1: x = 0. Let  > 0. Then, since xn → 0, there exists an M0 ∈ N such that ∀n ≥ M0 , xn = |xn − 0| < 2 . Choose M = M0 . Then, ∀n ≥ M , √ √ √ √ | xn − 0| = xn < 2 = . Case 2: x > 0. We have ∀n ∈ N, √ √ √ √ xn − x √ √ √ · ( xn + x) | xn − x| = √ xn + x 1 √ |xn − x| =√ xn + x 1 ≤ √ |xn − x|. x Hence, √ √ 1 0 ≤ | xn − x| ≤ √ |xn − x| x ∀n ∈ N. Hence, by the Squeeze Theorem, √ √ lim | xn − x| = 0. n→∞ Remark 11. Why must we do casework in the above proof ? Theorem 12 If {xn } is convergent and limn→∞ xn = x, then {|xn |} is convergent and limn→∞ |xn | = |x|. Proof : Firstly, note that ∀x ∈ R, √ x2 = |x|. Then, lim |xn | = lim n→∞ n→∞ √ p x2n = x2 = |x| by the previous theorem. Theorem 13 If c ∈ (0, 1), then limn→∞ cn = 0. If c > 1, then {cn } is unbounded. Proof : If 0 < c < 1, we claim that ∀n ∈ N, 0 < cn+1 < cn < 1. We can prove this through induction. Firstly, notice that 0 < c2 < c < 1 since c > 0 and c < 1. Now assume that 0 < cm+1 < cm . Then, multiply by c > 0 to obtain 0 < cm+1 · c = c(m+1)+1 < cm · c = c(m+1) . By induction, our claim holds. Thus, {cn } is a monotone decreasing sequence and is bounded below. Thus, {cn } is convergent. Let L = limn→∞ cn . We will prove that L = 0. Let  > 0. Then, ∃M ∈ N such that ∀n ≥ M , |cn − L| < (1 − c) 2 . Therefore, (1 − c)|L| = |L − cL| = |L − cM +1 + cM +1 − cL| ≤ |L − cM +1 | + c|cM − L|   < (1 − c) + c(1 − c) < (1 − c). 2 2 Therefore, ∀ > 0, |L| <  =⇒ L = 0. 4 Now let c > 1. We have to show that ∀B ≥ 0, ∃n ∈ N such that cn > B. Let B ≥ 0. Choose n ∈ N such that B n > c−1 . Then, cn = (1 + (1 − c))n ≥ 1 + n(c − 1) ≥ n(c − 1) > B. To see why this center inequality is true, see the last theorem shown in Lecture 1. 5