* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Worksheet 11 MATH 3283W Fall 2012 Basic definitions. A sequence (s
Survey
Document related concepts
Transcript
MATH 3283W Worksheet 11 Fall 2012 Basic definitions. A sequence (sn ) of real numbers is said to converge to s ∈ R if for any > 0, there exists a number N such that for any n ≥ N the inequality |sn − s| < holds. The number s is called the limit of sequence (sn ) and denoted by lim sn . If a sequence (sn ) does not converge to any number, it is said to diverge. 1. Prove the following statements using only the basic definitions. a) The sequence (an ) defined by an = n n2 +1 converges to a = 0. b) The sequence (bn ) defined by bn = 5n+1 7n−3 converges to b = 57 . c) The sequence (cn ) defined by cn = n3 −n n3 +2n d) The sequence (dn ) defined by dn = √ n n+100 e) The sequence (en ) defined by en = converges to c = 1. diverges. √ √ n + 1 − n converges to d = 0. f) The sequence (fn ) defined by fn = (−1)n diverges. g) The sequence {0.2, 0.22, 0.222, 0.2222, . . . } converges to g = 92 . h) The sequence (hn ) defined by hn = 0.99n converges to h = 0, but the sequence (jn ) with jn = 1.01n diverges. 2. Prove that if a sequence (sn ) converges to a strictly positive number s, then this sequence contains infinitely many positive terms. In practice, the following facts can be very useful. Theorem.1 Let (sn ) and (tn ) be convergent sequences with lim sn = s and lim tn = t. Then the following is true: a) lim(sn ± tn ) = s ± t b) lim(sn · tn ) = s · t c) lim stnn = st provided that tn 6= 0 for all n ∈ N and t 6= 0. d) for any k ∈ R, lim(k · sn ) = k · s and lim(k + sn ) = k + s Lemma(commonly known as the squeeze lemma or two policemen lemma). Let (an ) and (cn ) be convergent sequences such that lim an = lim cn = b. Then if (bn ) is such a sequence that an ≤ bn ≤ cn for all n, then (bn ) converes and lim bn = b. 3. For each of the given sequences, find its limit (if it exists). a) b) c) d) e) 1 1 + 3n+5 . an = 2n 2n−1 (n+3)10 bn = n10 +310 . 2) √ . cn = 10 sin(n n 2n +10 dn = √ . 3n −10 n3√ −1+100 en = n n+5 . Theorem 17.1, p.166. 2