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 &amp; N guarantees that j a n A j &lt; &quot;= 2 and j b n B j &lt; &quot;= 2 . Thus n &amp; N ) j ( a n b n ) ( A B ) j = j ( a n A ) + ( B b n ) j j a n A j + j B b n j &lt; &quot; 2 + &quot; 2 = &quot;: 4. Suppose A and B are nonempty sets of real numbers with A B: Then inf A &amp; inf B: Proof. Assume that A and B are bounded below. Then both sets have &amp;nite in&amp;ma. Let &amp; = inf A and = inf B: By de&amp;nition, is a lower bound for B: Take any a 2 A: Then a 2 B by the containment A B; so a: This shows that is a lower bound for A: Since &amp; is the greatest lower bound of A we can conclude &amp; &amp; : If B is not bounded below, then = 1 so the inequality &amp; &amp; is automatic. 1...
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