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370hw2solnsFall2011

# 370hw2solnsFall2011 - 2 Let N = max f N 1;N 2 g Then n&...

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MATH370 Fall 2011 Solutions to HW 2 Exercise 45. Prove that lim n !1 1 n 2 = 0 : Proof. Let " > 0 : By the Archimedean property of the reals, there exists a natural number N such that N > 1 =": Thus 0 < 1 =N < ": For every natural nuumber n , we have n N ) 1 n 2 ± 1 n ± 1 N < " ) 1 n 2 ² 0 < ": Thus 1 =n 2 ! 0 : Exercise 50. Let f a n g be a sequence of positive real numbers converging to A: Then A & 0 : Proof. Suppose A < 0 : Then ² A > 0 : We get a contradiction by setting " = ² A: limit, there exists a natural number N such that n N implies ² " < a n ² A < ": In particular, ² " < a N ² A < ": Adding A to both sides of the right hand inequality yields a N < A + " = 0 which contradicts the positivity of a N : This contradiction proves A & 0 : Exercise 55. Prove the limit law for di/erences of sequences. Proof. Suppose f a n g and f b n g are convergent sequences of real numbers with a n ! A and b n ! B: We must show that ( a n ² b n ) ! A ² B: Let " > 0 be given. Then there exist natural number N 1 and N 2 such that n N 1 implies j a n ² A j < "= 2 and n N 2 implies j b n ² B j < "=
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Unformatted text preview: 2 . Let N = max f N 1 ;N 2 g : Then n & N guarantees that j a n ² A j < "= 2 and j b n ² B j < "= 2 . Thus n & N ) j ( a n ² b n ) ² ( A ² B ) j = j ( a n ² A ) + ( B ² b n ) j ± j a n ² A j + j B ² b n j < " 2 + " 2 = ": 4. Suppose A and B are nonempty sets of real numbers with A ³ B: Then inf A & inf B: Proof. Assume that A and B are bounded below. Then both sets have &nite in&ma. Let & = inf A and ± = inf B: By de&nition, ± is a lower bound for B: Take any a 2 A: Then a 2 B by the containment A ³ B; so ± ± a: This shows that ± is a lower bound for A: Since & is the greatest lower bound of A we can conclude & & ±: If B is not bounded below, then ± = ²1 so the inequality & & ± is automatic. 1...
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