A First Course in Mathematical Analysis - download pdf or read online


ISBN-10: 0511348576

ISBN-13: 9780511348570

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N þ 1Þ , 2 ! 1þ n 2 and this final inequality certainly holds for n ! 3. 3 Prove that n23nþ2 < 1; for n > 2. More on inequalities We now look at a number of inequalities and methods for proving inequalities that will be useful later on. Àa þ bÁ2 Example 3 Prove that ab , for a, b 2 R. 2 Solution We tackle this inequality using the various rearrangement rules and a chain of equivalent inequalities until we obtain an inequality that we know must be true   aþb 2 a2 þ 2ab þ b2 ab , ab 2 4 , 4ab a2 þ 2ab þ b2 , 0 a2 À 2ab þ b2 , 0 ð a À bÞ 2 : This has the following geometric interpretation: The area of a rectangle with sides of length a and b is less than or equal to the area of a square with sides of length aþb 2 .

C) an ¼ (À 1)n, n ¼ 1, 2, . . Solution (a) The sequence {2n À 1} is monotonic because an ¼ 2n À 1 and anþ1 ¼ 2ðn þ 1Þ À 1 ¼ 2n þ 1; so that anþ1 À an ¼ ð2n þ 1Þ À ð2n À 1Þ ¼ 2 > 0; for n ¼ 1; 2; . : Thus {2n À 1} is increasing. ÈÉ (b) The sequence 1n is monotonic because 1 1 and anþ1 ¼ ; an ¼ n nþ1 so that In fact, strictly increasing. : nþ1 n ðn þ 1Þn ðn þ 1Þn ÈÉ Thus 1n is decreasing. Alternatively, since an > 0, for all n, and anþ1 n < 1; for n ¼ 1; 2; . ; ¼ nþ1 an it follows that anþ1 < an ; for n ¼ 1; 2; .

This sequence is null. For   ðÀ1Þn  1 jan j ¼  2  ¼ 2 ; n n so that, if we make the choice X ¼ p1ffiffi", it is certainly true that jan j < "; for all n > X: In terms of the " À X game, this simply means that whatever choice of " is made by player A, player B can always win by making the choice X ¼ p1ffiffin ! This is the case because, if n > p1ffiffi" ½¼ XŠ, then we have n2 > 1" or " > n12 ; hence, for n > X, we have jan j ¼ n12 < X12 ¼ ": Remarks n no Þ 1. In the sequence ðÀ1 , the signs of the terms made no difference to n2 whether the sequence was null, for they disappeared immediately we took the modulus of the terms in order to examine whether the definition (1) of a null sequence was satisfied.

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A First Course in Mathematical Analysis by DAVID ALEXANDER BRANNAN

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