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By Ekkehard Kopp

Construction at the easy suggestions via a cautious dialogue of covalence, (while adhering resolutely to sequences the place possible), the most a part of the booklet matters the imperative themes of continuity, differentiation and integration of actual capabilities. all through, the old context during which the topic was once constructed is highlighted and specific realization is paid to exhibiting how precision permits us to refine our geometric instinct. The goal is to stimulate the reader to mirror at the underlying options and ideas.

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Thus IXn - o] < B for all n > N, hence (x n ) con verges to ex as n ---+ 00. : The two previous results together imply: Every boundedmonotone sequence in IR is convergent. When these conditions are satisfied, limn x, equals SUPn x; when (xn) is increasing, and inf', X n when (xn) is decreasing. So we need to consider ways of identifying which sequences are monotone. Example I To decide whether a sequence (xn ) is monotone, we need to compare X n and X n+ l for every n E N. X n If the ratios remain less than 1 from some no onwards, it follows that Xn+l < X n for n ~ no.

N 1, less than (say)! for large enough n (take n ::: 21xl, using the Archimedes Principle). Hence this series converges also. This suggests that the series Ln ~ will converge for each x E ~, so that for each x we can take the sum of the series as the value of a certain function of x. But we need to wait a little longer to prove this plausible fact. 32 Analysis We saw in Further Exercise 5 on Chapter 2 that it has been known since the fourteenth century that Ln ~ converges to 2. What about Ln~? Here we == (n+l)2 X ~ == 1 (1 + 1)2 ---+ 1 hence the series converges On the have an+l an 2n+ 1 n2 2 n 2' .

In particular, this shows that we need only find two subsequences with different limits to prove that a sequence fails to converge: for example, (x n) with X n = (_I)n has subsequences (X2n) and (X2n-l) converging to 1 and -1 respectively; thus (x n) diverges. Completeness and Convergence 25 Example 6 A more subtle example is the following: let X n = sin(n) for all n E N. Now . (x ) > 1t an d si 71t < x < 6' l lz so nt -- 1,ml- 4 _ '12 wh en 61t < x < 2' sm (x ) < _ -'21 wh en 6" sm provides natural numbers such that sin(nl) > and sin(ml) < respectively.

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