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109_sp2011_ho_limits_sol

# 109_sp2011_ho_limits_sol - 109 Spring 2011 Limits and...

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109 Spring 2011 - Limits and Functions - Solutions Exercise. We say that a sequence f : Z + R is bounded if there is some number b R such that for each n Z + , | f ( n ) | ≤ b. (a) Prove that the harmonic sequence is bounded. Let b = 1. Then for each n Z + f ( n ) = 1 n 1 = b. (b) Prove that the sequence of (positive) powers of 2 is not bounded. We need to prove that for each b R , there is some n Z + such that f ( n ) > b . So let b be an arbitrary real number. If b 0, each positive power of 2 is greater than it. So, suppose b > 0. Let n b be the smallest integer bigger than or equal to b , n b = b . Then f ( n b ) = 2 n b 2 n > n. Exercise (Hard!) . Prove that if a sequence converges to a finite limit then it is bounded. To make the proof more readable, we prove below the (weaker) statement that if a sequence has positive values (that is, f : Z + R + ) and converges to a finite limit then it is bounded. The proof of the general case is very similar but a little longer.

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109_sp2011_ho_limits_sol - 109 Spring 2011 Limits and...

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