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0 ðby ‘completing the square’Þ , ðk À 1Þ2 À2 ! 0: This last inequality certainly holds for k ! 4, and so 2kþ1 ! (k þ 1)2 also holds for k ! 4. In other words: P(k) true for some k ! 4 ) P(k þ 1) true. It follows, by the Principle of Mathematical Induction, that 2n ! n2 , for & n ! 4. Problem 8 Prove that 4n > n4, for n ! 5. n n 2 n2 1 2 3 4 5 2 1 4 4 8 9 16 16 32 25 This assumption is just P(k). Since P(k þ 1) is: 2kþ1 ! (k þ 1)2. 1: Numbers 20 Three important inequalities in Analysis Our first inequality, called Bernoulli’s Inequality, will be of regular use in later chapters.

It follows that the sequence is not a null sequence. Note Notice that any positive value of " less than 1 will serve for our purpose here; there is nothing special about the number 12. These two examples illustrate the following strategy: Strategy for using the definition of null sequence 1. To show that {an} is null, solve the inequality janj < " to find a number X (generally depending on ") such that janj < ", for all n > X. 2. To show that {an} is not null, find ONE value of " for which there is NO number X such that janj < ", for all n > X.

The following example shows you how to do this: Example 1 Determine which of the following sets are bounded above, and which have a maximum element: (a) E1 ¼ [0, 2); (b) E2 ¼ f1n : n ¼ 1; 2; . g; (c) E3 ¼ N. Solution (a) The set E1 is bounded above. For example, M ¼ 2 is an upper bound of E1, since x 2; for all x 2 E1 : However, E1 has no maximum element. For each x in E1, we have x < 2, and so there is some real number y such that x < y < 2; by the Density Property of R. Hence y 2 E1, and so x cannot be a maximum element.

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