1873_solutions

# X is all we need even though most elementary calculus

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Unformatted text preview: ct that 2 1 j! 0e j0 e 21 ! we deduce that e  11 1 2 6 e which gives us e  3. d. Prove that if n is a positive integer we have n 1 j! 0e j0 3 n1 ! and deduce that n lim nÝ j0 1  e. j! The inequality n 0e j0 1 j! 3 n1 ! 1 j! e n1 ! follows at once from the inequality n 0e j0 and the fact that e  3. The fact that 259 n lim e nÝ 1 j! j0 0 now follows at once from the sandwich theorem. e. Prove that the number e is irrational. Solution: To obtain a contradiction, suppose that e is rational. Choose two positive integers p p and q such that e  q . Since n lim nÝ 1  e, j! j0 we have n lim nÝ q n jq1 1 j! nÝ lim 1 j! j0 q j0 1 j! e j0 1 j! and therefore, multiplying by q!, we obtain n lim nÝ Since both of the numbers q q! j0 j! jq1 p and q q! j! p q q!  q j0 q! j! q! are integers, we deduce that the number n lim nÝ q! j! jq1 must also be an integer. Now, on the one hand, the number n q! j! lim nÝ jq1 must be positive because n lim nÝ However, whenever j jq1 q 1 q! j! q!  1  0. q1 j!  jq1 q  1 we have q! 1  j! q1 q2  j  1 q1 jq and therefore n lim nÝ jq1 q! j! nq n nÝ lim jq1 1 q1 jq 1 1 nÝ lim q1 1 nÝ lim 1 q 1 1 q 1 j1 1 q1 j nq 1 q 1 We have therefore reached the impossible conclusion that the number n lim nÝ jq1 q! j! is an integer that lies between 0 and 1 and so we have reached the desired contradiction. 9. Prove that if x n is a sequence of positive numbers and if 260 n lim xx 1   n nÝ then lim n x n  . nÝ Deduce that lim n  e. n! nÝn Solution: We look first at the case   0. In order to show that lim n x n   nÝ we shall show that no number other than  can be a partial limit of the sequence n x n . Suppose that u is any number that is unequal to  and choose a number  between 0 and 1 such that the interval ,   does not contain the number u. We now use the facts that       and lim x n1   n Ý xn to choose a positive integer N such that the inequality n   xx 1   n  holds whenever n N. Now whenever n  N we have xn  xN x N1 xN x N2 x N1  xx n1 n and so x N     xn  xN nN nN which we can express as n xN N N   n xn  n N xN N .  Since lim n Ýn xN N N  n Ýn lim N xN 1 N x n can lie outside of the interval ,   n xn . and we conclude that u is not a partial limit of we deduce that no partial limit of the sequence n We now consider the case   0. In order to show that lim n x n  0 nÝ and, once again, we shall demonstrate that no number unequal to  can be a partial limit of the sequence n x n . Since it is already clear that a negative number can’t be a partial limit of n x n , we can confine our attention to positive numbers. Suppose that u  0 and choose a number  between 0 and 1 such that   u. We now use the fact that lim x n1  0 n Ý xn to choose an integer N such that the inequality 261 x n 1   xn holds whenever n N. Now whenever n  N we have xn  xN x N1 xN x N2 x N1  xx n1 n  xN...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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