PROBLEMS

# PROBLEMS - Exercise 2.5.7 Let{an{bn be two sequences of...

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Exercise 2.5.7 Let { a n } , { b n } be two sequences of positive terms. Prove that lim a n b n 6 lim a n lim b n , when the product on the right is not of the form 0 · ∞ . Proof : We distinguish 3 cases: lim a n = : Then by hypothesis lim b n > 0, so the right-hand side is equal to , and we have nothing to prove. By symmetry, the same happens when lim b n = . lim a n = 0 : Then by hypothesis lim b n < and the right-hand side is equal to zero. So, we must prove that lim a n b n = 0, or equivalently lim a n b n = 0 (indeed, if lim a n b n = 0 were true, then 0 6 lim a n b n 6 lim a n b n = 0 , so everything must equal zero). But this is true, since a n b n 6 Ma n , for all n N , where M is the upper bound of the sequence { b n } , which exists since lim b n < . So 0 6 a n b n 6 Ma n , for all n N , and by the squeeze theorem we get that lim a n b n = 0, as desired (again, we remind that lim a n = 0 holds since both lim a n and lim a n are equal to zero). By

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PROBLEMS - Exercise 2.5.7 Let{an{bn be two sequences of...

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