Introduction to Real Analysis

KðeÞ, we even have nk ! ok ! KðeÞ in order that jxnk À xj < e. consequently the subsequence ðxnk Þ additionally converges to x. Q. E. D. three. four. three Examples (a) limðbn Þ ¼ zero if zero < b < 1. now we have already obvious, in instance three. 1. 11(b), that if zero < b < 1 and if xn :¼ bn , then it follows from Bernoulli’s Inequality that limðxn Þ ¼ zero. on the other hand, we see that considering the fact that C03 12/08/2010 11:35:19 web page seventy nine three. four SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM seventy nine zero < b < 1, then xnþ1 ¼ bnþ1 < bn ¼ xn in order that the series ðxn Þ is reducing. it's also transparent that zero xn 1, so it follows from the Monotone Convergence Theorem three. three. 2 that the series is convergent. enable x :¼ lim xn . due to the fact ðx2n Þ is a subsequence of ðxn Þ it follows from Theorem three. four. 2 that x ¼ limðx2n Þ. in addition, it follows from the relation x2n ¼ b2n ¼ ðbn Þ2 ¼ x2n and Theorem three. 2. three that À Á2 x ¼ limðx2n Þ ¼ limðxn Þ ¼ x2 : for this reason we should have both x ¼ zero or x ¼ 1. because the series ðxn Þ is lowering and bounded above by means of b < 1, we deduce that x ¼ zero. (b) limðc1=n Þ ¼ 1 for c > 1. This restrict has been acquired in instance three. 1. 11(c) for c > zero, utilizing a slightly inventive argument. We supply right here another process for the case c > 1. be aware that if zn :¼ c1=n ; then zn > 1 and znþ1 < zn for all n 2 N. (Why? ) hence through the Monotone Convergence Theorem, the restrict z :¼ limðzn Þ exists. by way of Theorem three. four. 2, it follows that z ¼ limðz2n Þ. furthermore, it follows from the relation z2n ¼ c1=2n ¼ ðc1=n Þ1=2 ¼ z1=2 n and Theorem three. 2. 10 that À Á1=2 ¼ z1=2 : z ¼ limðz2n Þ ¼ limðzn Þ for that reason now we have z2 ¼ z whence it follows that both z ¼ zero or z ¼ 1. considering zn > 1 for all n 2 N, we deduce that z ¼ 1. & We depart it as an workout to the reader to contemplate the case zero < c < 1. the subsequent result's according to a cautious negation of the definition of limðxn Þ ¼ x. It results in a handy approach to determine the divergence of a chain. three. four. four Theorem permit X ¼ ðxn Þ be a chain of actual numbers, Then the subsequent are similar: (i) The series X ¼ ðxn Þ doesn't converge to x 2 R. (ii) There exists an e0 > zero such that for any okay 2 N, there exists nk 2 N such that nk ! okay and jxnk À xj ! e0 . (iii) There exists an e0 > zero and a subsequence X zero ¼ ðxnk Þ of X such that jxnk À xj ! e0 for all okay 2 N. evidence. ðiÞ ) ðiiÞ If ðxn Þ doesn't converge to x, then for a few e0 > zero it truly is very unlikely to discover a average quantity ok such that for all n ! ok the phrases xn fulfill jxn À xj < e0. that's, for every okay 2 N it's not real that for all n ! ok the inequality jxn À xj < e0 holds. In different phrases, for every ok 2 N there exists a traditional quantity nk ! ok such that jxnk À xj ! e0 . ðiiÞ ) ðiiiÞ enable e0 be as in (ii) and allow n1 2 N be such that n1 ! 1 and jxn1 À xj ! e0 . Now enable n2 2 N be such that n2 > n1 and jxn2 À xj ! e0 ; permit n3 2 N be such that n3 > n2 and jxn3 À xj ! e0 . proceed during this strategy to receive a subsequence X zero ¼ ðxnk Þ of X such that jxnk À xj ! e0 for all ok 2 N. ðiiiÞ ) ðiÞ feel X ¼ ðxn Þ has a subsequence X zero ¼ ðxnk Þ fulfilling the situation in (iii). Then X can't converge to x; for if it did, then, by way of Theorem three.