Advanced calculus : an introduction to mathematical analysis - download pdf or read online

By S. Zaidman.

ISBN-10: 9810227043

ISBN-13: 9789810227043

Ch. 1. Numbers --
ch. 2. Sequences of genuine numbers --
ch. three. limitless numerical sequence --
ch. four. non-stop features --
ch. five. Derivatives --
ch. 6. Convex services --
ch. 7. Metric areas --
ch. eight. Integration.

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Show that it is a convergent sequence. Acn, We shall see that it is a Cauchy sequence, let us evaluate \an — a m |, assume m > n and put p = m — n; we have | ^ n — Gn+pl < |^n ~ &n+l \ + | a n + l ~ «-n+2| + . . 4- | a n + p _ ! - an+p\ < A(cn + c n + 1 + . . + c 7 1 ^" 1 ) 1- c 1- c as c n —► 0, we get Ve > 0, j4^c n < e if n > n(e) hence \an - a n + p | < e for n > n(e), p G N. Example 11. Let us take ax < a 2 and define an by the recursion a n + 2 = \{an + a n + i ) . Prove that it is a convergent sequence.

It follows that (an + 6n) —► +oo. In fact, take any M > 0, we have 6n > M for n > n, take also e = 1, we get a n > a — 1 for n > l\ thus, for n > max (n,I) we obtain an + bn > M -ha — 1. Finally, if Mi > 0 is given, take M such that M + a - 1 > Mx. Example 10. Let (a n )i° be a sequence, such that \an — a n + i | < Vn G N with some c e (0,1). Show that it is a convergent sequence. Acn, We shall see that it is a Cauchy sequence, let us evaluate \an — a m |, assume m > n and put p = m — n; we have | ^ n — Gn+pl < |^n ~ &n+l \ + | a n + l ~ «-n+2| + .

Thus let us assume that: lim xn = £i, and lim xn — £2- We have accordingly the inequalities: \xn — £\\ < e for n > ri\ and \xn - £ri2 (where e > 0 is in M). ) we obtain \£i - £i\ < \£i - xn\ + \xn - £2\ < 2e (We G E, e > 0). This inequality is not always possible (unless £\ — £2 = 0). In fact, if 29 30 Advanced Calculus \£i - h\ > 0 and e is taken = \\£i - £2\ we obtain \£x - £2\ < \\£i - £2\ which is obviously false. □ We say also that: (x n )J° is divergent, if $£ G E such that limx n = £.

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