A First Course in Analysis by Donald Yau

By Donald Yau

This e-book is an introductory textual content on genuine research for undergraduate scholars. The prerequisite for this booklet is an exceptional history in freshman calculus in a single variable. The meant viewers of this booklet comprises undergraduate arithmetic majors and scholars from different disciplines who use genuine research. because this e-book is geared toward scholars who do not need a lot earlier adventure with proofs, the speed is slower in past chapters than in later chapters. There are countless numbers of routines, and tricks for a few of them are incorporated.

Readership: Undergraduates and graduate scholars in research.

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B) Is {an } necessarily a Cauchy sequence if C = 1? (16) Let {an } be a bounded sequence with α = inf{an } and β = sup{an }. (a) If α =/ an for any n, prove that there exists a decreasing subsequence of {an } that converges to α. (b) If β =/ an for any n, prove that there exists an increasing subsequence of {an } that converges to β. (17) In each case, construct a sequence {an } with the given properties. (a) (b) (c) (d) There exist subsequences converging to 1, 2, and 3. For every positive integer M , there exists a subsequence converging to M .

Since lim an = L there exists a positive integer N1 such that n ≥ N1 implies ∣an − L∣ < . 5in analysis-yau Sequences 43 Likewise, there exists a positive integer N2 such that implies ∣bn − L∣ < . 3 Thus, for n ≥ N = max{N1 , N2 }, we have n ≥ N2 ∣cn − an ∣ ≤ ∣bn − an ∣ ≤ ∣bn − L∣ + ∣L − an ∣ < 3 + 3 = 2 . 3 So for n ≥ N = max{N1 , N2 }, we have ∣cn − L∣ ≤ ∣cn − an ∣ + ∣an − L∣ < 2 + = . 3 3 This shows that lim cn = L. 14. The sequence with an = − cos n n satisfies 1 1 ≤ an ≤ n n because −1 ≤ cos x ≤ 1 for every real number x.

Then for integers n, m ≥ N 1 1 1 , }≤ < . n+1 m+1 N +1 This shows that {an } is a Cauchy sequence. 10 the sequence {an } is convergent. 19. Consider the sequence with 1 1 1 an = 1 + 2 + 2 + ⋯ + 2 . 2 3 n As in the previous example, we want to show that this is a Cauchy sequence. For m > n we have 1 1 1 am − an = + +⋯+ 2 (n + 1)2 (n + 2)2 m 1 1 1 < + +⋯+ n(n + 1) (n + 1)(n + 2) (m − 1)m 1 1 1 1 1 1 )+( − ) + ⋯( − ) =( − n n+1 n+1 n+2 m−1 m 1 1 1 < . = − n m n Given > 0 choose a positive integer N > 1 .

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