By S. Zaidman.

Ch. 1. Numbers --

ch. 2. Sequences of genuine numbers --

ch. three. limitless numerical sequence --

ch. four. non-stop services --

ch. five. Derivatives --

ch. 6. Convex capabilities --

ch. 7. Metric areas --

ch. eight. Integration.

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Ft. Sequences of Real Numbers 45 Next, we note the following: VA; G N, 3nk > k, and ank > {3 — e. (In fact, otherwise, 3k0 G N such that Vn > fc0, we get an < f3 — e\ this gives bko+i < ft — £ < Pi contrary to relation f3 = inf bn). Choose now e = 1; we have an < f3 + 1 for n > ft, and 3fli > 1, with a n i > (3 — 1. For 7ii=max (n,ni) we get therefore )S — 1 < o n i < /? + 1, and ni > 1 . , hence (3 e S. In a similar vay we see that a = liminf a n also belongs to 5. Thus, the (a)-part is verified.

Xn+1, hence (1 — x)s = 1 — xn+1). Let us now note a slight extension of Theorem 1. Theorem 3. A monotone increasing sequence is either convergent (to a real number) or is convergent to +oo. Proof. Let (xn) be a monotone increasing sequence. The case when it is upper bounded is covered by Theorem 1. Suppose then that it is not bounded from above. Therefore, for any M > 0, there is n 0 G N such that ano > M. It follows that an > M if n > n 0 . Hence, in this case, lim an = 4-oo. Example 1. Consider the sequence: xn = (1 4- ^ ; ) n .

Let (a n ), (bn) be two sequences, where an -^ a and 6 n —> +oo. 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| + .