319HW5Correction

319HW5Correction - MAT 319/320 Correction of HW5 Exercise 1...

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Unformatted text preview: MAT 319/320 Correction of HW5 Exercise 1. Page 80, #1. Proof. Take for example the following sequence: x 2 k = k,x 2 k +1 = 1 . square Exercise 2. Page 80, #3. Proof. Recall that f n satisfies the relation f n +2 = f n +1 + f n and that all the f n are > . therefore we deduce that f n +2 f n +1 = 1 + f n f n +1 .We know that x n = f n +1 f n has a limit L . Notice that L cannot be zero (because then f n f n +1 would then be unbounded, which is not the case because it is equal to f n +2 f n +1- 1 , which converges to L- 1 ). Thus we can apply the quotient theorem and obtain the equality L = 1 + 1 L . This implies that L 2 = L + 1 .By solving this equation and keeping the positive root we get L = 1 + 5 √ 2 . square Exercise 3. Page 80, #8a. Proof. Let’s compare x n +1 = (3( n +1)) 1/2( n +1) and x n = (3 n ) 1/2 n . We have x n +1 2 n. 2( n +1) = 3( n + 1) 2 n = (3 n ) 2 n . (1 + 1 n ) 2 n , whereas x n 2 n. 2( n +1) = (3 n ) 2( n +1) = (3 n ) 2 n . (3 n ) 2 , therefore after some integer K , the sequence is decreasing (because (1 + 1 n ) 2 n is eventually strictly smaller than...
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319HW5Correction - MAT 319/320 Correction of HW5 Exercise 1...

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