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By Ekkehard Kopp

Construction at the uncomplicated suggestions via a cautious dialogue of covalence, (while adhering resolutely to sequences the place possible), the most a part of the e-book matters the significant themes of continuity, differentiation and integration of actual capabilities. all through, the historic context during which the topic was once constructed is highlighted and specific cognizance is paid to displaying how precision permits us to refine our geometric instinct. The goal is to stimulate the reader to mirror at the underlying techniques and ideas.

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So the subsequence found in • Theorem 2-Bolzano-Weierstrass Theorem - - - Every bounded monotone sequence in IR has a convergent subsequence. PROOF Given the sequence (x n ) choose a monotone subsequence (x n, ) . Since [x, : n E N}is bounded, so is its subset {x n, : r EN}. Hence (xn, ) is a bounded monotone sequence, and therefore converges, by the remark following Theorem 1. This seemingly innocent result will have far-reaching consequences when we consider more general subsets of ~ and functions ~ I~ IR.

2. Example 2 We return to power series, and apply Theorem 1. In particular, for [x] < 1, the power series Ln2:0 x" is absolutely convergent, and its Cauchy product with itself is Lk>O(k + l)xk. 1 + ... 1 = k + 1, as there are k + 1 terms in the above definition of Ck. On the other hand L:o x" = l~x' so we have shown that Ln2:1 nxn- l = Lk2:0(k + l)xk = (l~x)2 whenever Ixl < 1. ) 46 Analysis Example 3 Similarly, as Lm2:0 ~ converges absolutely for all x E ~, we can explore the Cau~ product (L:=o ~)(L:o for general x, y E ~.

Is the sum of finitely many of the an, so 1m :::; s for every mEN. Since the bk ~ 0 this means that (t m ) is an increasing sequence, bounded above by s, and hence converges to SUPm 1m = 1 :::; s. On the other hand, we can exchange the roles of the two series: now if Lk bk is a series with positive terms and sum t, then it follows that the rearrangement Ln an of its terms also converges and has sum s :::: 1. Hence s = 1. Now consider an E IR more generally: suppose that the real series Ln an is absolutely convergent, that Ln lanl has sum S, and that Lk bk is a rearrangement of Ln an· Then Lk Ibkl is a rearrangement of the terms of Ln lanl.

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