By S. Zaidman.

Ch. 1. Numbers --

ch. 2. Sequences of actual numbers --

ch. three. limitless numerical sequence --

ch. four. non-stop services --

ch. five. Derivatives --

ch. 6. Convex services --

ch. 7. Metric areas --

ch. eight. Integration.

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Then limsup a n and liminf an both belong to S and they are the largest (respectively the smallest) element of S. Proof. (a) Let us prove for instance that limsup an G S. Let be (3 = limsup anTherefore /3 = lim bn where bn = sup ( a n , a n + i , . . ) . Then, given any e > 0, we find bn < (3 + e for n > n, hence a n < 6 n < /3 + e for n. > 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).

We shall now discuss again some typical examples of convergent sequences. 5. (1) The sequence ( ( | ) n ) is a null sequence, in R. 1, we know the lower estimate: 3 n > n V n G N. 5, Proposition 5). (2) We explain now an extension of the previous example: lim an = 0 for all a G R, 0 < a < 1. 32 Advanced Calculus In fact, if 0 < a < 1, then £ > 1, so £ = 1 + ft, where ft > 0; accordingly we can write that a = ~ ^ , ft > 0. 1), it is in fact true in the present situation: for 7i = 1 it is an obvious equality.

5 Limsup and Liminf of a Bounded Sequence As easily seen, a bounded sequence is not necessarily convergent. However, as indicated in B-W theorem, it has convergent subsequences. Actually, there can be various subsequences with different limits. This situa tion is best handled by means of the concept of limit superior and limit inferior of any bounded sequence. Let therefore (a n ) nG N be an arbitrary bounded sequence of real numbers. For each n, let An be the set { a n , a n + i , a n + 2 , .