Download Advanced calculus : an introduction to mathematical analysis by S. Zaidman. PDF

By S. Zaidman.

Ch. 1. Numbers --
ch. 2. Sequences of actual numbers --
ch. three. endless numerical sequence --
ch. four. non-stop services --
ch. five. Derivatives --
ch. 6. Convex services --
ch. 7. Metric areas --
ch. eight. Integration.

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Example text

Hence, as an —► 0, we find, given any e > 0, a number rio(e) G N, such that I ^2 apbp\ < e for m> n> n0(e). n+l I D Corollary, (the alternating series test) Let (an) be a monotone decreasing CO sequence of positive numbers and let lim an = 0. TAen £/ie series ^2(—l)n_1an l is convergent. The proof is obvious. 4 Summability CO Consider a divergent series: Ylan- Therefore, the sequence of finite sums: (sn) I where sn = a\ 4- a2 4- • • • an is not convergent. It can however happen that the sequence of arithmetic means: (tn) where 1, tn = —(«1 + S 2 H Sn) n is a convergent sequence.

The root test). Suppose an > 0 for all n. If lim sup n v/a^ CO CO l l < 1, then the series 53 an converges. If lim sup nyfa~^ > 1 £/ien t/ie series 53 a n diverges. Proof. If lim sup 7\fan~ = q < 1 we take q', q < q' < 1 and we find no € N such that y ^ < q' for n > no- Hence an < {q')n for n > no, and therefore CO 53 &n is convergent. l If lim sup ny/a^ = r > 1 we take r', 1 < r1 < r. 4 (II) (2). Take ni > 1, such that n\fa~^[ > r'. Then, take n 2 > ni, such that n \fa^ > r', continuing this way we find a subsequence (a,nk)kLi of (a n ) such 54 Advanced Calculus that aUk > r'nk.

There are four cases, according as A is bounded above (or not) and bounded below (or not). 1. A bounded below but not bounded above: Let a — inf A. We will show that A = (a, +oo) or A — [a,+oo). First, notice that A C [a, oo). Let then r > a. There exists x G A, a < x < r. Also, as A is not bounded above there exists y £ A, y > r. Hence x < r < y where x, y G A. By convexity => r G A. Hence, (a, + oo) C A C [a, oo). 2. A bounded above but not below; a similar reasoning gives that, with b = sup A (-00,6) C A C (-00,6].

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