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Unformatted text preview: Math 3110 Spring ’09 Homework 2: Selected Solutions [1.2  1 (e)]: Claim : b n = a n 2 √ 2 n + 1 = ∑ n k =1 1 √ k 2 √ 2 n + 1 is a strictly negative increasing sequence. Therefore b n has a limit. Proof 1 First notice that a n > R n +1 1 dx √ x = 2 √ n + 1 2 and a n < R n +1 dx √ x = 2 √ n + 1, which immediately yields b n = a n 2 √ n + 1 < (2 √ n + 1 2) 2 √ n + 1 > 2 and b n = a n 2 √ n + 1 < 0 for all n . Moreover, b n is increasing: b n +1 b n = a n + 1 √ n + 1 2 √ n + 2 ( a n 2 √ n + 1) = 1 √ n + 1 2 √ n + 2 + 2 √ n + 1 = 1 2 p ( n + 2)( n + 1) + 2( n + 1) √ n + 1 Clearly ‡ 1 2 p ( n + 2)( n + 1) + 2( n + 1) √ n + 1 · > ⇒ b n +1 b n > 0, and we observe: 1 2 p ( n + 2)( n + 1) + 2( n + 1) = 3 + 2 n 2 p n 2 + 3 n + 2 > 3 + 2 n 2 √ 3 n 2 (for n > 2) = 3 + (2 √ 3) n > (since 2 √ 3 > 0) So b n is an increasing negative sequence (bounded above by zero 2 ); hence it has a limit. 1 I was initially unable to insert figures into this document, so I’ve given a somewhat less geometric proof than the one modeled on that of Prop. 1.5B in your text. 2 Tighter upper bounds than zero can be obtained with a little extra work. If you are interested, see the continuation of this problem on pg. 3. 1 [1.6 3]: Claim : a n +1 = 2 a 2 n with a ∈ (0 , 1 2 ) is strictly decreasing and bounded below....
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This note was uploaded on 11/09/2009 for the course MATH 3110 taught by Professor Ramakrishna during the Fall '08 term at Cornell.
 Fall '08
 RAMAKRISHNA
 Math

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