By S. Zaidman.

Ch. 1. Numbers --

ch. 2. Sequences of actual numbers --

ch. three. endless 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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**Extra info for Advanced calculus : an introduction to mathematical analysis**

**Example text**

Thus, for A; > max (711,712) we get L — e < ak < L + e. We next establish the special situation of limsup and liminf of a bounded sequence with respect to all possible limits of its subsequences. We have pre cisely the following Theorem 3. Let (a n ) n 6 N be a bounded sequence in E. Define S = {x G E, 3(a nfc ), subsequence of (a n ) nG N, such that ank —► x}. 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.

See below) This inequality in fact implies that, for n > 2 'n2-l\ n = A n * 1 - —z \ n J Then consider the quotient ^ 1 , 1 > l - n - - jz = l - - = n n = (gf^r = ^ S ^ - ^ n-1 . • T h u s xn > *n-i for n = 2 , 3 , . . (Here we used the "strict" Bernoulli's inequality: If x > — 1 and x / 0 , then (1 + x)n > 1 + nx for all n > 2: again, it is true for n = 2; assume it true for n = m; thus (1 + x ) m > 1 -f rare. Multiply by (1 4- x) which is > 0. We obtain: (1 + x ) m + 1 > (1 + mx)(l + x) = 1 + (m + l)a + mx 2 > 1 4- (m + l):r if r r / 0 ) .

Take n\ > n, such that a n i is not a peak; therefore, 3n2 > ni, such that an2 > ani. Again, aU2 is not a peak, thus, 3ns > n2, such tht an3 > a n2 ; and as we continue the same way, we obtain the subsequence ani, an2, an3,... such that ani < an2 < an3 . . < . . (an increasing sequence). Next, let us examine the second possibility. Take n\ G N, ani is a peak point. Then, 3n2 > ni, an2 is also a peak point; 3^3 > ri2, anz is also a peak point and so on. We obtain a subsequence of peak points, ani, an2, an3 ...