74-hw10s - MATH 74 HOMEWORK 10 - DUE MONDAY, APRIL 30 1. We...

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MATH 74 HOMEWORK 10 - DUE MONDAY, APRIL 30 1. We have 30 3 = 27 , 000 so that 10 4 30 3 < 10 5 and thus a 3 = 4. Similarly 30 4 = 810 , 000 so that 10 5 30 4 < 10 6 and thus a 4 = 5. If p N 0 , the set of numbers x satisfying 10 p x < 10 p +1 is precisely the set of all p + 1-digit numbers. We thus see that a n is the number of digits in 30 n minus one. 2. Fix n N . By definition of the sequence we have 10 a n 30 n . Applying L and using properties from last week’s homework we conclude a n L (10) nL (30); using the fact that L (10) = 1 and dividing by n we conclude L (30) a n n . Similarly, since 30 n < 10 a n +1 we conclude nL (30) < ( a n + 1) L (10) and thus L (30) < a n +1 n . 3(a). Yes, b n L (30) for all n . 3(b). No. We computed a 3 = 4 and a 4 = 5, so that b 3 = 4 3 and b 4 = 5 4 . Since b 3 > b 4 the sequence is not increasing. 3(c). Yes, c n L (30) for all n . 3(d).
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74-hw10s - MATH 74 HOMEWORK 10 - DUE MONDAY, APRIL 30 1. We...

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