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Question: hi i am not understanding how they did this proof...

Question details

Hi, I am not understanding how they did this proof. I am totally lost. If you could explain in more detail the logic which they used to derive their conclusion I would be most grateful. Thanks.

The book is Bartle, 4th edition, Introduction to Real Analysis.

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Background info

2.1.13 Examples (a) Let a 2 0 and b20. Then We consider the case where a > 0 and b 〉 0, leaving the case a = 0 to the reader. It follows from 2. 1.5(i) that + b 0. Since b2-a2-(b-a) (b + a), it follows from 2.1.7(c) that b-a > 0 implies that b--r > 0. Also, it follows from 2.1.10 that b--a2 > 0 implies that b - a>0. Ка 〉 0 and b 〉 0, then Ja 〉 0 and b 〉 0. Since = (Ja)-and b = (VE) 2, the second implication is a consequence of the first one when a and b are replaced by va and vb, respectively. We also leave it to the reader to show that if a > 0 and b > 0, then

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(b) Let Z() be the sequence of real numbers defined by z11, zn+1 :- v22n for nEN. We will show that lim(2n)-2. Note that zı = 1 and z2 = /2; hence l < zl < Z2 < 2. We claim that the sequence Z is increasing and bounded above by 2. To show this we will show, by Induction, that I 〈 ZnくZn+1く2 for all n E N. This fact has been verified for n l. Suppose that it is true for n - k; then 2 < 2zk < 2zk+ 4, whence it follows (why?) that [In this last step we have used Example 2.1.13(a).] Hence the validity of the inequality 1くZkくZk+1く2 İmplíes the validity of 1くZk+1くZk+2 〈2. Therefore I 〈Zn〈 -n+1 〈 2 for all n E N Since Z- (2n) is a bounded increasing sequence, it follows from the Monotone Convergence Theorem that it converges to a number zsup. It may be shown directly that supízn) -2, so that z-2. Alternatively we may use the method employed in part (a). The relation n+i- V22n gives a relation between the nth term of the 1-tail Z1 of Z and the nth term of Z. By Theorem 3.1.9, we have lim Z1zlim Z. Moreover, by Theorems 3.2.3 and 3.2.10, it follows that the limit z must satisfy the relation Hence must satisfy the equation 72-27, which has the roots ž 0, 2. Since the terms of -= (z.) all satisfy I 〈Zn-2, it follows from Theorem 3.2.6 that we must have 1 〈 z 〈 2. Therefore ž = 2

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