# HW#6 - a n = 0 since | a |< 1 Thus lim(1 a a n = 1 1-a(c...

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SOLUTIONS TO HOMEWORK # 6 9.8 (a) Let M > 0 and take N = M 1 / 3 . Then n > N n 3 > M . Thus by deﬁnition lim n 3 = . (b) lim - n 3 = - lim n 3 = -∞ , using the result of (a). (c) If n is even ( - n ) n > 1. If n is odd ( - n ) n < 0. Thus the limit does not exist. (d) lim(1 . 01) n = by example 9.7 in the text. (e) Let M > 0 and let N = M . Then certainly n > N n n > n > M . Thus by deﬁnition lim n n = . 9.9 (a) Fix M > 0. By deﬁnition of lim s n = + , N : n > N s n > M . Now take N 1 = max { N,N 0 } . Then n > N 1 t n s n > M . Thus by deﬁnition lim t n = . (b) For M > 0. By deﬁnition of lim t n = -∞ , N : n > N t n < - M . Now take N 1 = max { N,N 0 } . Then n > N 1 s n t n < - M . Thus by deﬁnition lim s n = -∞ . 9.18 (a) It will suﬃce to show (1 - a )(1 + a + ... + a n ) = 1 - a n +1 . (1 - a )(1 + a + ... + a n ) = (1 + a + ... + a n ) - ( a + a 2 + ... + a n +1 ) . Cancelling appropriate terms gives that this is just 1 - a n +1 . (b) From (a), lim(1 + a + ... + a n ) = lim ± 1 - a n +1 1 - a . Using the limit theorems, lim ± 1 - a n +1 1 - a = 1 - a lim a n 1 - a . Example 9.7 gives lim

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Unformatted text preview: a n = 0, since | a | < 1. Thus lim(1 + a + ... + a n ) = 1 1-a . (c) This is application of part (b) with a = 1 3 . Thus lim ± 1 + 1 3 + 1 2 + ... + 1 3 n ¶ = 1 1-1 2 = 3 2 . 1 (d) Let M > 0 and choose N = M . Then n > N ⇒ (1+ a + a 2 + ... + a n ) ≥ 1+1+ 1 2 + ... +1 n = n +1 > M , thus by deﬁnition if a ≥ 1 then lim(1+ a + ... + a n ) = ∞ . PROBLEM A (1) lim ‡ a n · 3 = a 3 lim 1 n 3 = ( a 3 )(0) = 0. (2) lim ‡ n a · 3 = 1 a 3 lim n 3 = ∞ . (3) lim ‡ a n · 1 /n = lim a 1 /n lim n 1 /n = 1 1 = 1 by Examples 9.7 (c,d). (4) lim ‡ n a · 1 /n = lim n 1 /n lim a 1 /n = 1 1 = 1. (5) We have 0 < a b < 1 so it remains to use Example 9.7(b) in the Text. 2...
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## This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Penn State.

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HW#6 - a n = 0 since | a |< 1 Thus lim(1 a a n = 1 1-a(c...

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