Hwk2_Solns

# Hwk2_Solns - Math 3110 Spring 09 Homework 2 Selected...

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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 0 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 · > 0 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 > 0 (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

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[1.6 -3]: Claim : a n +1 = 2 a 2 n with a 0 (0 , 1 2 ) is strictly decreasing and bounded below. Proof There are two quick ways to prove this; one is by direct computation and the other is by induction. (1) Direct Computation : Notice that a 1 = 2 a 2 0 , a 2 = 8 a 4 0 , ..., a n = 2 2 n +1 a 2 n 0 . So clearly a n > 0 when a 0 (0 , 1 2 ). Moreover a n +1 a n = 2 2 n +2 a 2 n +1 0 2 2
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