By John Srdjan Petrovic

ISBN-10: 1351381431

ISBN-13: 9781351381437

ISBN-10: 1466565640

ISBN-13: 9781466565647

ISBN-10: 2592642692

ISBN-13: 9782592642691

Sequences and Their Limits Computing the LimitsDefinition of the restrict houses of Limits Monotone Sequences The quantity e Cauchy Sequences restrict more suitable and restrict Inferior Computing the Limits-Part II genuine Numbers The Axioms of the Set R effects of the Completeness Axiom Bolzano-Weierstrass Theorem a few strategies approximately RContinuity Computing Limits of capabilities A overview of capabilities non-stop capabilities: ARead more...

summary: Sequences and Their Limits Computing the LimitsDefinition of the restrict homes of Limits Monotone Sequences The quantity e Cauchy Sequences restrict improved and restrict Inferior Computing the Limits-Part II genuine Numbers The Axioms of the Set R results of the Completeness Axiom Bolzano-Weierstrass Theorem a few techniques approximately RContinuity Computing Limits of services A evaluate of capabilities non-stop capabilities: a geometrical standpoint Limits of features different Limits homes of continuing features The Continuity of ordinary services Uniform Continuity homes of constant capabilities

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**Additional info for Advanced Calculus : Theory and Practice**

**Sample text**

First, the sequence an is increasing so, if n ≥ N , an ≥ aN > L − ε. 1) On the other hand, L is an upper bound for A, so an ≤ L for all n ∈ N. Thus, if n ≥ N , L − ε < an ≤ L < L + ε. 2) We conclude that |an − L| < ε, and the theorem is proved. One of the disadvantages of the deductive method is that every assertion needs to be proved. We are not going to aim that high, and we will be occasionally guilty of taking some facts as obvious. Sometimes, though, the omission will be only temporary. One such issue appeared in Chapter 1, and it concerns the definition of the floor function.

P − a b √ Therefore, there exists a positive integer q between p a2 and p √ √ 2 2 p −˜ a (if a ˜ = 0 we can just take N = 1), and hence a ˜ + N > 0. If we now denote a = a ˜ + N , b = ˜b + N , then 0 < a < b, so the previous proof applies and we get r, ρ ∈ (a, b).

Proof. First we will show that an is an increasing sequence. Instead of an+1 − an > 0, we will establish that an+1 /an > 1. ) We notice that n+1 n+1 1 1 1+ 1 + 1 an+1 n+1 n+1 1+ = . = n an n 1 1 1+ 1+ n n 18 Advanced Calculus: Theory and Practice Further, 1 n+1 = 1 1+ n 1+ n+2 n+1 n+1 n = n2 + 2n 1 n(n + 2) = =1− 2 . (n + 1)2 n2 + 2n + 1 n + 2n + 1 Therefore, using Bernoulli’s Inequality, n+1 1 1 + 1 1 n n+1 ≥ 1 − (n + 1) · 2 =1− = . n + 2n + 1 n+1 n+1 1 1+ n It follows that n an+1 ≥ an n+1 1+ 1 n = n n+1 = 1, n+1 n and the sequence {an } is increasing.

### Advanced Calculus : Theory and Practice by John Srdjan Petrovic

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