57323-0136147054_10

# 57323-0136147054_10 - Section 10.2 10.2.1 The most obvious...

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Section 10.2 10.2.1: The most obvious pattern is that a n = n 2 for n = 1. 10.2.2: The most obvious pattern is that a n =5 n 3for n = 1. 10.2.3: The most obvious pattern is that a n = 1 3 n for n = 1. 10.2.4: The most obvious pattern is that a n = ( 1) n 1 2 n 1 = ± 1 2 ² n 1 for n = 1. 10.2.5: The most obvious pattern is that a n = 1 3 n 1 for n = 1. 10.2.6: The most obvious pattern is that a n = 1 n 2 +1 for n = 1. 10.2.7: Perhaps the most obvious pattern is that a n =1+( 1) n for n = 1. 10.2.8: One pattern is that a n = 15 2 5 2 · ( 1) n . Another is that a n · ± 1+ 1 ( 1) n 2 ² for n = 1 . 10.2.9: lim n →∞ 2 n 5 n 3 = lim n →∞ 2 5 3 n = 2 5 0 = 2 5 . 10.2.10: lim n →∞ 1 n 2 2+3 n 2 = lim n →∞ 1 n 2 1 2 n 2 +3 = 0 1 0+3 = 1 3 . 10.2.11: lim n →∞ n 2 n +7 2 n 3 + n 2 = lim n →∞ 1 n 1 n 2 + 7 n 3 2+ 1 n = 0+0+0 2+0 =0 . 10.2.12: This sequence diverges because a n = n 3 10 n 2 > n 3 10 n 2 +10 n 2 = n 20 + as n + . 10.2.13: Example 9 tells us that if | r | < 1, then r n 0as n + .T ak e r = 9 10 to deduce that lim n →∞ ³ ( 9 10 ) n ´ =1+0=1 . 1083

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10.2.14: Example 9 tells us that if | r | < 1, then r n 0as n + .T ak e r = 1 2 to deduce that lim n →∞ ± 2 ( 1 2 ) n ² =2 0=2 . 10.2.15: Given: a n =1+( 1) n for n = 1. If n is odd then a n 1) = 0; if n is even then a n = 1 + 1 = 2. Therefore the sequence { a n } diverges. To prove this, we appeal to the deFnition of limit of a sequence given in Section 10.2. Suppose that { a n } converges to the number L .Le t f = 1 2 and suppose that N is a positive integer. Case 1: L = 1. Then choose n = N such that n is odd. Then a n =0,so | a n L | = | 0 L | = L = 1 >f. Case 2: L< 1. Then choose n = N such that n is even. Then a n =2,so | a n L | = | 2 L | L> 1 >f . No matter what the value of L , it cannot be made to Ft the deFnition of the limit of the sequence { a n } . Therefore the sequence { a n } = { 1+( 1) n } has no limit. (We can’t even say that it approaches + or −∞ ;itdoesnot .) 10.2.16: Because 1 + ( 1) n =0if n is odd and 2 if n is even, 0 5 1) n n 5 2 n for all n = 1. Therefore, by the squeeze law for sequences, lim n →∞ 1) n n =0. 10.2.17: We use l’Hˆ opital’s rule for sequences (Eq. (9) of Section 10.2): lim n →∞ a n = lim n →∞ 1) n n ( 3 2 ) n = lim x →∞ 1 ± x 1 / 2 ( 3 2 ) x = ± lim x →∞ 1 2 x 1 / 2 ( 3 2 ) x ln ( 3 2 ) =0 . 10.2.18: We use the squeeze law for sequences (Theorem 3 of Section 10.2): 1 5 sin n 5 1 for all integers n = 0, and therefore 1 3 n 5 sin n 3 n 5 1 3 n for all integers n = 1. By the result in Example 9 of Section 10.2, 1 / 3 n n + . Therefore lim n →∞ sin n 3 n . 10.2.19: ±irst we need a lemma. 1084
Lemma: If r> 0, then lim n →∞ 1 n r =0. Proof: Suppose that 0. Given f> 0, let N =1+[ [1 /f 1 /r ]] . T h e n N is a positive integer, and if n>N , then n> 1 /f 1 /r ,s oth a t n r > 1 /f . Therefore 1 n r 0 <f .

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57323-0136147054_10 - Section 10.2 10.2.1 The most obvious...

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