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By de Boor C. (ed.)

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20(4), (2y)n/n! -+ 0 as n -+ and therefore we can find an Nsuch that, for any n ~N,yn/n! ] Show that the product of the limits of the two sequences «(1 + xn -I) n) and «(1 -xn-I)n) is equal to 1. 9 Theorem Every sequence has a monotone subsequence. Proof Let (xn ) be any sequence of real numbers. We must construct a . subsequence (xn r ) which is either increasing or decreasing. We distinguish two . cases. (i) Every set {xn: n > N} has a maximum. In this case we call find a sequence ( nr) of natural numbers such that = maxxn n>1 and so on.

E. I = m. 23 Theorem Suppose that (i) Ifxn~a(n=I,2, (ii) If Xn ~ b (n = 1,2, Xn -+ I as n -+ 00. ),then/~a. ), then I ~ b. , ~ b (n = 1,2, ... , ~ - b (n = 1,2, ... ). Hence we need only prove (i). Let E > O. ; < I + E. e. 1- 38 Convergent sequences But Xn ~ a (n = 1,2, ... ) and so, for any n > N, a ~ Xn < 1 + e. Hence, given any e > 0, a < 1 + €. 7 it follows that a < I. ; > a (n = 1,2, ... ), it is tempting to conclude that I > a. But this may not be true. For example, Iln -+ as n -+ and 11n > (n = 1,2, ...

5, that Xn ~ va = 1, 2, ... ). Hence (n IXn+1 - va I <; = 2va (XI 2va val2n We know that, if I y I < 1, then y n --+ 0 as n --+ that y2n --+ 0 as n --+ Hence our argument shows that Xn --+ va as n 00. 2 it follows 00. XI - va / --+ 00 provided that <1. 2va We have already assumed that XI ~ va and so we have shown that Xn --+ va as n --+ provided that va <; X I < 3 va. 5 that this was true under the weaker hypothesis that XI> O. /a. /2. /2 < 2 (because 12 = 1 < 2 and 22 = 4> 2). For a better estimate tab a = 2 and Xl = 2 in the argument above.

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