Chapter_2

# Chapter_2 - II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 X 8 1 II.1.1 1 2 3 P L K Establish the following(a lim n!1 n n 1 = 1(c

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Unformatted text preview: II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 X 8 1 II.1.1 1 2 3 P L K Establish the following: (a) lim n !1 n n +1 = 1 (c) lim n !1 2 n p +5 n +1 n p +3 n +1 = 2 ; p > 1 (b) lim n !1 n n 2 +1 = 0 (d) lim n !1 z n n ! = 0 ; z 2 C : Solution (a) lim n !1 n n + 1 = lim n !1 1 1 + 1 =n = 1 : (b) lim n !1 n n 2 + 1 = lim n !1 1 =n 1 + 1 =n 2 = 0 : (c) lim n !1 2 n p + 5 n + 1 n p + 3 n + 1 = lim n !1 2 + 5 n 1 & p + n & p 1 + 3 n 1 & p + n & p = 2 : (d) Let m be a positive integer so that m > 2 j z j and let n > m , then & & & & & z n n ! & & & & = j z j m j z j n & m m ! ¡ ( m + 1) ¡ ( m + 2) ¡ ¡¡¡ ¡ n & & j z j m m ! ¡ j z j m + 1 ¡ j z j m + 2 ¡ ¡¡¡ ¡ j z j n & j z j m m ! ¡ j z j n = K n ; where K = j z j m +1 m ! and because K n ! as n ! 1 , it follows by inst&ngning that j z j n n ! ! as n ! 1 as was to be shown. 2 II.1.2 1 2 3 P L K For which values of z is the sequence f z n g 1 n =1 bounded? For which values of z does the sequence converge to ? Solution The sequence is bounded for j z j & 1 , and the series converge to for j z j < 1 . 3 II.1.3 1 2 3 P L K Show that f n n z n g converges only for z = 0 . Solution If we choose n > 2 j z j , then 2 < n j z j , 2 n < n n j z j n , & & n n z n & & for j n n z n j & & & & & n n z n & & & & & 2 n ! + 1 ; for z 6 = 0 . 4 II.1.4 1 2 3 P L K Show that lim n !1 N ! N k ( N & k )! = 1 ; k & . Solution N ! N k ( N & k )! = N ( N & 1) ::: ( N & k +1) N ¡ N ¡ ::: ¡ N = & 1 ¡ 1 N ¡& 1 ¡ 2 N ¡ ::: & 1 ¡ k & 1 N ¡ ! 1 as N ! 1 ( k is &xed) 5 II.1.5 1 2 3 P L K Show that the sequence b n = 1 + 1 2 + 1 3 + &&& + 1 n ¡ log n; n ¢ 1 ; is decreasing, while the sequence a n = b n ¡ 1 =n is increasing. Show that the sequences both converge to the same limit & . Show that 1 2 < & < 3 5 . Remark. The limit of the sequence is called Euler&s constant. It is not known whether Euler&s constant is a rational number or an irrational number. Solution We text we have that b n = 1 + 1 2 + 1 3 + &&& + 1 n ¡ 1 + 1 n ¡ log n; a n = 1 + 1 2 + 1 3 + &&& + 1 n ¡ 1 ¡ log n: We trivially have that log n = Z n 1 1 t dt (1) : An elementary estimation gives 1 k + 1 < k +1 Z k 1 t dt < 1 k (2) : We take the di/erence a n +1 ¡ a n and by (1) and (2) we have that a n +1 ¡ a n = 1 n ¡ log ( n + 1) + log n = 1 n ¡ n +1 Z n dt t > ; thus a n +1 > a n , so the sequence a n is increasing. We take the di/erence b n +1 ¡ b n and by (1) and (2) we have that 6 b n +1 & b n = 1 n + 1 & log ( n + 1) + log n = 1 n + 1 & n +1 Z n dt t < ; thus b n +1 > b n , so the sequence b n is decreasing. We have after some investigation on the increasing sequence a n that a n ¡ a 7 = 49 20 & log 7 > 1 2 ; n ¡ 7 ; thus 1 2 < a n , if n ¡ 7 ....
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## This note was uploaded on 11/13/2010 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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Chapter_2 - II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 X 8 1 II.1.1 1 2 3 P L K Establish the following(a lim n!1 n n 1 = 1(c

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