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Unformatted text preview: a n = 0, since  a  < 1. Thus lim(1 + a + ... + a n ) = 1 1a . (c) This is application of part (b) with a = 1 3 . Thus lim ± 1 + 1 3 + 1 2 + ... + 1 3 n ¶ = 1 11 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.
 Spring '08
 OLSAVSKYGREGORM

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