Chapter 4 Sequences Section 4.2 Limit Theorems Suppose that ( - PowerPoint Presentation

Chapter 4 Sequences

Section 4.2Limit Theorems

Suppose that (s n) and (tn ) are convergent sequences with lim s n = s and lim tn = t. Then To simplify our work with convergent sequences, we prove several useful theorems in this section. The first theorem shows that algebraic operations are compatible with taking limits. Theorem 4.2.1 (a) lim (sn + tn) = s + t (c) lim (sn tn) = s t (d) lim (sn /tn) = s / t, provided that tn  0 for all n and t  0. Proof: |(sn + tn) – (s + t)| = |(sn – s) + (tn – t)| (a) For all n  we have  |sn – s| + |tn – t | by the triangle inequality. Given any  > 0, since sn  s, there exists N1  such that |sn – s| <  /2 for all n  N1. Likewise, there exists N 2  such that |tn – t | <  /2 for all n  N2. Now let N = max {N1, N2}. Then for all n  N we have |(sn + tn) – (s + t)|  |sn – s| + |tn – t | Thus, lim (sn + tn) = s + t. The proof of (b) is Exercise 4. (b) lim ( ksn) = ks and lim (k + sn) = k + s, for any k  .

Letting M = max { M 1 , | t |}, we obtain the inequality Thus there exists M 1 > 0 such that | sn |  M1 for all n. Suppose that (sn ) and (tn) are convergent sequences with lim sn = s and lim tn = t. ThenTheorem 4.2.1(c) lim (sn t n) = s t Proof:We know from Theorem 4.1.13 that the convergent sequence (sn) is bounded. Now let N = max {N1 , N2}. Then n  N implies that Thus, lim sn tn= s t .Given any  > 0, there exist natural numbers N1 and N2 such that | tn – t | <  /(2M) when n ≥ N1 and | sn – s | <  /(2M) when n ≥ N 2. This time we use the inequality| sn tn – st | = |(sn t n – sn t) + ( sn t – st)| |(sn tn – s n t)| + |( s n t – st)|= | sn |  | t n – t | + |t |  |s n – s |

Suppose that (s n) and (tn ) are convergent sequences with lim s n = s and lim tn = t. Then Theorem 4.2.1 Proof: Since sn /tn = sn(1/tn), it suffices from part (c) to show that lim (1/t n) = 1/t. That is, given  > 0, we must make (d) lim (sn /tn) = s / t, provided that tn  0 for all n and t  0. for all n sufficiently large. To get a lower bound on how small the denominator can be, we note that, since t  0, there exists N1  such that n ≥ N1 implies that | t n – t | < | t | /2. Thus for n  N1 we have by Exercise 3.2.6(a). There also exists N2  such that n  N2 implies that Let N = max {N1, N2}. Then n  N implies thatHence lim (1/ t n)= 1/ t . 

Example 4.2.2* Show that This is Example 4.1.9* from the last section. We have s n = 0 5 Now lim (1/n) = 0, so lim (1/n2) = 02 = 0, lim (– 6 /n2) = (– 6)(0) = 0,and lim [5 – (6 /n2)] = 5. 08Likewise, lim (3/n) = 0,so lim [8 – (3/ n)] = 8. And,

Suppose that ( sn) and ( t n ) are convergent sequences with lim sn = s and lim tn = t. If sn  tn for all n  , then s  t. Then  = (s – t)/2 > 0, and we have 2 = s – t and t +  = s – .Thus there exists N1  such that n ≥  N1 implies that s –  < sn < s + . Theorem 4.2.4 Proof: Suppose that s > t.t  s Similarly, there exists N2  such that n ≥ N2 implies that t –  < tn < t + .Let N = max {N1, N2 }. Then for all n ≥ N we have tn < t +  = s –  < sn,This contradicts the assumption that sn  tn for all n, and we we conclude that s  t.  Corollary 4.2.5 If ( tn) converges to t and tn  0 for all n  , then t  0. Proof: Exercise 4(b). 

Then since ( sn + 1/sn) converges to L, there exists N  such that n  N implies that A “ratio test” for convergenceTheorem 4.2.7 Suppose that (sn) is a sequence of positive terms and that the sequence of ratios (sn + 1 / sn) converges to L. If L < 1, then lim sn = 0. Proof: Corollary 4.2.5 implies L  0. Suppose L < 1. Let  = c – L so that  > 0. Let k = N + 1. Then for all n  k we have n – 1  N, so that It follows that, for all n  k,Letting M = sk /ck, we obtain 0 < sn < M cn for all n  k. Since 0 < c < 1, Exercise 4.1.7(f ) implies that lim cn = 0.Thus lim sn = 0 by Theorem 4.1.8. Then there exists a real number c such that 0  L < c < 1.

To make 3n /2 > M, we want n > 2 M /3.To handle a sequence such as (sn) = (n) where the terms get larger and larger, we introduce infinite limits. A sequence (sn ) is said to diverge to + , and we write lim sn = +  if for every M  there exists a natural number N such that n  N implies that s n > M.Show thatDefinition 4.2.9A sequence (sn) is said to diverge to – , and we write lim s n = –  if for every M  there exists a natural number N such that n  N implies that sn < M . Example 4.2.11 This time we need a lower bound on the numerator.We find that 4n 2 – 3  4 n2 – n2 = 3n2, when n  2.For an upper bound on the denominator, we have n + 2  n + n = 2n, when n  2. Thus for n  2 we obtainSo given any M  , take N > max {2, 2 M / 3}.The formal proof is in the text.

We conclude with two theorems for infinite limits. The proofs are left for the exercises. Suppose that ( s n) and (tn) are sequences such that s n  t n for all n  .(a) If lim sn = + , then lim tn = + .(b) If lim tn = – , then lim sn = – .Let (sn) be a sequence of positive numbers. Then lim sn = +  if and only if lim (1 / sn) = 0. Theorem 4.2.12Theorem 4.2.13