Download Problems in Mathematical Analysis 1: Real Numbers, Sequences by W. J. Kaczor, M. T. Nowak PDF

By W. J. Kaczor, M. T. Nowak

We study by means of doing. We research arithmetic by means of doing difficulties. This publication is the 1st quantity of a chain of books of difficulties in mathematical research. it's typically meant for college students learning the fundamental ideas of study. besides the fact that, given its association, point, and choice of difficulties, it will even be an excellent selection for academic or problem-solving seminars, really these aimed toward the Putnam examination. the quantity can be compatible for self-study.

Each element of the publication starts with quite uncomplicated routines, but can also comprise really hard difficulties. quite often a number of consecutive workouts are thinking about diversified points of 1 mathematical challenge or theorem. This presentation of fabric is designed to aid pupil comprehension and to inspire them to invite their very own questions and to begin learn. the gathering of difficulties within the publication can be meant to aid academics who desire to include the issues into lectures. suggestions for all of the difficulties are supplied.

The e-book covers 3 subject matters: genuine numbers, sequences, and sequence, and is split into elements: workouts and/or difficulties, and strategies. particular subject matters coated during this quantity comprise the subsequent: easy homes of actual numbers, persevered fractions, monotonic sequences, limits of sequences, Stolz's theorem, summation of sequence, checks for convergence, double sequence, association of sequence, Cauchy product, and countless items

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Additional resources for Problems in Mathematical Analysis 1: Real Numbers, Sequences and Series

Example text

In case that the sequence is not convergent to any real number, we say that it is divergent. 2. Let a be some real number and xn : Then lim xn a . a for all n È N. 3. Investigate whether the following limits exist. 5 2 Fig. 12: Ôn, Ô¡1Õn Ôn 1ÕßnÕ for n 1 to n 50 and asymptotes. 50 40 30 20 10 10 20 30 Fig. 13: Ôn, Ôn2 1ÕßnÕ for n 1 to n 38 40 50 n 50 and an asymptote. 2) . 3) Solution: Fig. 11, Fig. 12 and Fig. 3 don’t exist. Indeed lim n n 1 n 1. 4) For the proof, let ε be some real number 0. Further, let n0 be some natural number 1ßε.

2Ù . Since 0 È S, S is a non-empty. Further, S does not contain real numbers y 2 since the last inequality implies that y 2 ¡ 2 Ôy ¡ 2ÕÔy 2Õ 2 2. Hence S is bounded from above. We define x : sup S. In the following, we prove that x2 2 by excluding that x2 2 and that x2 2. First, we assume that x2 2. Then it follows for n È N¦ that ¢ x 1 n ª2 ¡2 x2 ¡ 2 Hence if n x2 ¡ 2 2x 1 . n 2x n n12 x2 ¡ 2 2x n n1 (2x 1)ß(2 ¡ x2 ) it follows that ¢ x 1 n ª2 2 and therefore that x (1ßn) È S. As a consequence, x is no upper bound for S.

2nÕ 1ß2 for all n È N¦ ÞØ1Ù is convergent. Solution: The sequence x1 , x2 , . . is bounded from below by 0. In addition, xn 1 2n 1 xn 2Ôn 1Õ xn for all n È N¦ and hence x1 , x2 , . . is decreasing. Hence x1 , x2 , . . 13. See Fig 15. 15. Let S be a non-empty subset of R. We say that S is bounded from above (bounded from below) if there is M È R such that x M (x M) for all x È S. 16. Let S be a non-empty subset of R which is bounded from above (bounded from below). Then there is a least upper bound (largest lower bound) of S which will be called the supremum of S (infimum of S) and denoted by sup S (inf S).

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