HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!


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1 Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following conjectural identity F n x = xn n! For convenience, let us define G n x = xn n! as F n x = G n x. For n 0 and x > 0 we note that G nx = xn n! = xn n! = = = + xn n! x + n =. for n 0 and x > 0, so that can be stated = xn n! n = = G n x. 2 We proceed by induction on n to prove, i.e. F n x = G n x. For n = 0, we immediately verify that G 0 x = ln x = F 0 x. For the inductive step, we assume that is true for n = m. This means that F m x = G m x, while 2 says that G m x = G m+x. Now we calculate F m+ x = 0 F m tdt = 0 G m tdt = lim r G m tdt = lim r G m+tdt = lim G m+ x G m+ r by the Fundamental Theorem of Calculus = G m+ x lim G m+ r In order to complete the inductive step, it remains to show that lim G m+ r = 0. This can be seen by r m writing G m+ r = m +! ln r m+ =. Note that as r goes to 0, the fisrt factor clearly converges to 0 /r whereas the second factor also converges to 0 by L Hopital s rule.
2 Math 7 Fall 205 Now says that n!f n = following inequalities: = = =. Since fx = x n = + n + Hence we have that + ln n ln n =2 n + = n!f n ln n x = =2 n+ x is strictly decreasing for x > 0, we obtain the n x = + = + ln n, x = lnn + > ln n. < ln n =. Since lim + ln n =, we deduce that ln n n ln n n!f n lim =. n ln n Problem A2 Let a, a 2, and b, b 2, be sequences of positive real numbers such that a = b = and b n = b n a n 2 for n = 2, 3,. Assume that the sequence b j is bounded. Prove that converges, and evaluate S. S = a a n n= The relation b n = b n a n 2 for n = 2, 3, can be written as a n = b n + 2 b n. Then for n = 2, 3,, = a a n a a 2 a n = a b b 2 b n b 2 + 2b b n + 2 = 3 b b 2 b n b + 2 b 2 + 2b b n + 2 since a = b = 3b b 2 b n = b + 2b b n + 2 = 3 b b 2 b n b n b + 2b b n + 2 b b 2 b n b n b + 2b b n + 2 = 3 b b 2 b n 2 b + 2b b n + 2 b b 2 b n b n b + 2b b n + 2 b b 2 b n b n For convenience, set c n = for n =, 2,, so that the above identity can be b + 2b b n + 2 written as = 3 a a n 2 c n c n for n = 2, 3,. Then for every positive integer m, we have n= a a 2 a n = a + n=2 a a 2 a n = c n c n = c c m = 3 2 c m, 3 n=2
3 Math 7 Fall 205 where the last identity follows by the fact that c = 3. Note that c m is clearly positive by construction since b, b 2, are all positive. We also now that the sequence b j is bounded by some number M > 0. Now observe that 0 < c m = b b + 2 b 2 b b m b m + 2 = + 2/b + 2/b 2 < + 2/b m + 2/M m. Since + 2/M converges to 3 2. >, this implies that lim m c m = 0. Thus we deduce from 3 that S = n= a a n Problem A3 Let, 2, 3,, 2005, 2006, 2007, 2009, 202, 206, be a sequence defined by x = for =, 2,, 2006 and x + = x + x 2005 for Show that the sequence has 2005 consecutive terms each divisible by Define a sequence y by y = 0 for =, 2,, 2005, y 2006 = and y + = y + y 2005 for Then we can easily compute y 2006 = y 2007 = = y 40 =, y 402 = 2, y 403 = 3,, y 607 = In particular, we have that y 40 = x, y 402 = x 2,, y 607 = x Now an easy strong induction shows that x = y +400 for =, 2,. Note that y is clearly a sequence of integers, so Useful Facts 2 from Lecture 5 see the remar below says that the sequence is periodic modulo In particular, since y, y 2,, y 2005 are all zero and therefore divisible by 2006, there exist infinitely many integers such that y +, y +2,, y are all divisible by Choose such a with 40, then x 4009 = y +, x 4008 = y +2,, x 2005 = y are all divisible by Remar. For the sae of completeness, we state and prove Useful Facts 2 from Lecture 5. Lemma. If x n is a sequence of integers satisfying a linear recurrence x n = a x n +a 2 x n 2 + +a x n, then x n is a periodic sequence in modulo m for every integer m. Proof. Without loss of generality, we may assume that a 0. Consider the vectors of the form x n, x n+,, x n+. Note that there are only finitely many such vectors modulo m, so there exist integers p and N such that x N+p, x N++p,, x N+ +p x N, x N+,, x N+ mod m. 4
4 Math 7 Fall 205 We claim that x n+p x n mod m for all n. It suffices to prove that the following holds for all n: x n+p, x n++p,, x n+ +p x n, x n+,, x n+ mod m. 5 We proceed by the twoway induction on n. The case when n = N is given by 4. For the inductive step, we assume that 5 holds for n = r. We wish to prove that 5 holds for n = r and n = r +. For n = r +, it suffices to prove that x r++p x r+ mod m. This can be seen by x r++p = a x r+ +p + a 2 x r+ 2+p + + a x r+p a x r+ + a 2 x r a x r mod m by recursion by the induction hypothesis = x r+ by recursion For n = r, it suffices to prove that x r +p x r mod m. This can be seen by x r +p = a x r+ +p a x r+ 2+p a x r+p by recursion a x r+ a x r+ 2 a x r mod m by the induction hypothesis = x r by recursion Problem B3 Let x 0 = and for n 0, let x n+ = 3x n + x n 5. In particular, x = 5, x 2 = 26, x 3 = 36, x 4 = 72. Find a closedform expression for x a means the largest integer a. Note that x n is an integer for n = 0,, 2,. By definition of the floor function, we have the following inequality for n 0: x n 5 < xn 5 xn 5. 6 Then the recursion x n+ = 3x n + x n 5 gives an inequality 3 + 5xn < x n x n. Multiplying by 3 5 yields 4x n 3 5 < 3 5x n+ 4x n. 7
5 Math 7 Fall 205 Now observe that for n =, 2,, x n+ 6x n = 3x n + x n 5 3 5xn < 4x n < 4x n +, x n+ 6x n = 3x n + x n 5 > 3 5xn 4x n where in each line, the first and the second inequality respectively follow from 6 and 7. Hence we have that 4x n < x n+ 6x n < 4x n + for n =, 2,. Since x n+ 6x n is an integer, the only possibility is that x n+ 6x n = 4x n. In other words, we have the following recurrence for n : x n+ = 6x n 4x n. 8 The characteristic polynomial x 2 6x + 4 has two roots 3 ± 5, so there exist constants c and d such that x n = c3 + 5 n + d3 5 n for n = 0,, 2,. Substituiting n = 0 and n =, we obtain a system of equations = c + d and 5 = c d3 5, which can be solved by c = , d = Hence 8 yields the following expression of x 2007 : 0 0 x 2007 =
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