By Horst R. Beyer

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**Example text**

2n ¡ 1Õ xn : 2 ¤ 4 . . Ô2nÕ 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).

Let x1 , x2 , . . be a sequence of elements of R and x È R. , from the n0 -th member on, all remaining members of the sequence are within a distance from x which is less than ε. 1 In this case, we say that the 1 As a consequence, only finitely many members have distance 36 ε from x. 25 10 20 30 Fig. 11: Ôn, Ôn 1ÕßnÕ for n 40 1 to n 50 n 50 and asymptotes. 0 sequence x1 , x2 , . . is convergent to x. Note that in this case for ε xn xn ¡ x x xn ¡ x x ε x for all n È N¦ , apart from finitely many members of the sequence, and hence that x1 , x2 , .

Then it follows that xk È Ik0 , xl È Ik0 and therefore that xk ¡ xl aß2k0 ε . Hence xn1 , xn2 , . . 10. For the following, the Bolzano-Weierstrass theorem will be fundamental. 57. 12. Let x1 , x2 , . . , for which there is M 0 such that xn M for all n È N. Then x1 , x2 , . . is convergent. Proof. Since x1 , x2 , . . is increasing and bounded from above, it follows that this sequence is also bounded. , a sequence xn1 , xn2 , . . that corresponds 46 to a strictly increasing sequence n1 , n2 , .