By S. Zaidman.

Ch. 1. Numbers --

ch. 2. Sequences of actual numbers --

ch. three. endless numerical sequence --

ch. four. non-stop capabilities --

ch. five. Derivatives --

ch. 6. Convex services --

ch. 7. Metric areas --

ch. eight. Integration.

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**Extra resources for Advanced calculus : an introduction to mathematical analysis**

**Example text**

Then (z ) = ((1 + ; r b ) n + 2 ) i s a subsequence of ((1 4- ^ ) m ) , and accordingly lim zn n — e. Obviously lim (1 4- r r o ) 2 = 1. ). Example 3. We examine the sequence (xn) where xn — (1 4- J ) n . )*-H)"(^)*-K)"(^r(^rFrom this relation we easily find that L = lim (1 + J ) n = e • e • 1 = e 2 as guessed previously. Example 4. Let us define a sequence (an) in the following way: ai = \/2, a 2 = V 2 + \/2, a 3 = Y 2 + V 2 + v ^ , . . , a n + i = ^ 2 + an . We shall see that this is a bounded increasing sequence of real numbers.

13. True or false? (explain): if (x n ) and {xnyn) are bounded sequences in Q, then (y n ) is also a bounded sequence (in Q). 14. Let (a n ), (6 n ) be sequences in Q and \an\ < bn\fn G N. Show that if (6 n ) is a null-sequence then (a n ) is also a null sequence. 15. Prove that the sequence (a n ) where an = | ( 1 — ^ ) Vn G N is a Cauchy sequence. 16. ) is order preserving: p > q in <2 => /(p) > /(g) in R. 17. Let x be a fixed real number. If x < e for all e G R, e > 0, prove that x < 0. 18. ).

As well as the obvious limits: ^ —> 0(n —* oo), Vp > 1. It results: l i m EM. = M n_oo Q(n) bo* (ii) If * > *, we have: %& = n / c - ^ » + | + - g = n *-* . ^ Now obviously lim nk~£ = +oo, lim a n = f11 Next we have Proposition. Let (A n ), (/xn) 6e sequences in R, and An —► oo, /xn —> \i. Then if \i > 0, An • fin —■> oo; if fi < 0, An • // n —► —oo. In fact, in first case we have: VA > 0, 3 n G N, such that An > A for n > n. Also /x — e < fj,n < (1 + e for n > n i , and with e = ^, jin > |/x, n > n2.