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Unformatted text preview: Math 3110 Homework 2 Solutions Exercise 3.1.2 Claim. If a n is a nonnegative sequence, lim a n = 0 = lim a n = 0 . Proof. Let a n be a nonnegative sequence. Assume lim a n = 0 (this means that for any > 0, there is a natural number N such that for all n N ,  a n  < ). Let > 0 and set = 2 . Since lim a n = 0, then there exists a natural number N such that for all n N ,  a n  < = 2 , which implies that a n < . That for every > 0, such an N exists, means that lim a n = 0. Exercise 3.2.5 Let { a n } be a convergent sequence of integers, with a n L . We prove that { a n } is eventually constant, that is, a n = L for large n . Since a n L , then there is some natural number some N such that  a n L  < 1 4 for all n N . This requires a n to lie in the interval ( L 1 4 ,L + 1 4 ) for all n N . As this interval has length 1 2 , it can contain at most 1 2 = 1 integer, say, M . Thus, a n = M for all n N , which is to say, { a n } is eventually constant.is eventually constant....
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 '08
 RAMAKRISHNA
 Math

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