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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 X X X X X X X X X X X X X X X X 2 X X X X X X 3 X X X X X 4 X X X X X 5 X X X X X X X X 6 7 8 1 II.1.1 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 m j z j m j z j m & & j z j m m ! 1 2 1 2 1 2 = j z j m m ! 1 2 n & m ! as n ! 1 , it follows by inst&ngning that & & z n n ! & & ! as n ! 1 as was to be shown. 2 II.1.2 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 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 and j n n z n j = j n & z j n > 2 n ! 1 when n ! 1 for z 6 = 0 . 4 II.1.4 Show that lim N !1 N ! N k ( N & k )! = 1 ; k & . Solution We have that 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 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 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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 Winter '07
 Vis

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