Homework_Set_3_Solution

# Homework_Set_3_Solution - Homework 3 Solutions 1 Apostol...

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Unformatted text preview: Homework 3 Solutions October 12, 2005 1 Apostol page 45, problem 12: a) 1 k ! k- 1 Y r =0 n- r n = 1 k ! n ( n- 1) ... ( n- ( k- 1)) n.n.....n = n ! k !( n- k )! 1 n k Therefore 1 + n X k =1 1 k ! k- 1 Y r =0 1- r n ! = 1 + n X k =1 n ! k !( n- k )! 1 n k = 1 + n ! 1 n + n ! 2!( n- 2)! 1 n 2 + ... + 1 n n = 1 + 1 n n . b) 1 + 1 n n > 1 + n ! 1 n ≥ 2 . Now 1- r n ≤ 1 so k- 1 Y r =0 1- r n < 1 and therefore 1 + 1 n n < 1 + n X k =1 1 k ! . Also 1 k ! ≤ 1 2 k- 1 1 for all k ≥ 1, so n X k =1 1 k ! < n X j =0 1 2 j < 2 Therefore 1 + n X k =1 1 k ! < 3 2 Let k = n + 1: lim k →∞ ( k- 1) 3 k 3 = k 3- 3 k 2 + 3 k- 1 k 3 = 1- 3 k + 3 k 2- 1 k 3 Now we know that 1 k → 0 as k → ∞ and therefore- 3 k → 0 as k → ∞ . Using comparison of sequences with 1 k , we see that 3 k 2 and 1 k 3 also → 0 as k → ∞ . Therefore, (1- 3 k + 3 k 2- 1 k 3 ) → 1 as k → ∞ . And so lim n →∞ n 3 ( n + 1) 3 = 1 ....
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Homework_Set_3_Solution - Homework 3 Solutions 1 Apostol...

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