By Sterling K. Berberian

**Uploader's Note:** Ripped from SpringerLink.

The booklet deals an initiation into mathematical reasoning, and into the mathematician's way of thinking and reflexes. particularly, the basic operations of calculus--differentiation and integration of features and the summation of limitless series--are equipped, with logical continuity (i.e., "rigor"), ranging from the genuine quantity procedure. the 1st bankruptcy units down designated axioms for the true quantity process, from which all else is derived utilizing the logical instruments summarized in an Appendix. The dialogue of the "fundamental theorem of calculus," the point of interest of the ebook, specially thorough. The concluding bankruptcy establishes an important beachhead within the thought of the Lebesgue vital by way of hassle-free potential.

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**Sample text**

Exercises 1. Let (an) be a sequence in lR. Prove that (an) is unbounded if and only if there exists a subsequence (a nk ) such that lank I ~ k for all k. 2. Let (an) be a sequence in lR . Prove that the following conditions are equivalent: (a) (an) is divergent; (b) for every a E lR there exist an E > 0 and a subsequence (a nk ) such that lank - al ~ E for all k. 3. *True or false (explain): The sequence an subsequence. * = sin n has a convergent 4. True or false (explain): If (an) is any sequence in lR, then the sequence has a convergent subsequence.

9. Theorem. If A c IR is nonempty and bounded above, and if M = sup A, then there exists a sequence (xn) in A such that Xn ~ M. 5). <:; < Similarly, if A c IR is nonempty and bounded below, then inf A is the limit of a sequence in A. Exercises 1. If an ~ a, bn 2. Suppose an ~ ~ b and an ~ a. Define 1 Sn Prove that Sn ~ bn frequently, show that a = -n Lak. n k=l a. } ~ b . 3. Sequences 42 3. Show that the sequence is convergent. } 4. Let a and b be fixed real numbers and define a sequence (xn) as follows: Xl = a , X2 = b and, for n > 2, Xn is the average of the preceding two terms, that is, Xn = ~(Xn-2 + xn-d.

From [an+l' bn+lJ c [an, bnJ we see that of the 24 2. First Properties of lR it follows that the sequence (an) is increasing and bounded above (for example by bl ), whereas (b n ) is decreasing and bounded below (for example by al) . 3) . 6 (and its 'dual', with arrows reversed) we have -bn i -b , so an + (-b n ) i a + (-b) , therefore bn - an 1 b - a . Since bn - an 2: 0 for all n , it follows that b - a 2: 0 ; then an::; a ::; b ::; bn , so [a, b] c [an, bn ] for all n, therefore n 00 [a, b] c [an, bn ].