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The first five terms of the sequence B*, listing the partial sums of B+, are B* = 0/1, 1/2, 7/6, 23/12, 163/60, As we continue to calculate partial sums, we keep adding values that get closer and closer to 1 The terms in B* grow at an ever-increasing rate If we choose any positive real number, no matter how large, we can eventually generate an element of B* that exceeds it Therefore, the sequence B* of partial sums does not converge 6 Here s the series again, expressed in summation notation: S+ =
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When we write out the first five terms of S+ as fractions, we get the sum S+ = 1/10 + 1/100 + 1/1000 + 1/10,000 + 1/100,000 + Expressing the terms as powers of 10, we have S+ = 10 1 + 10 2 + 10 3 + 10 4 + 10 5 + Expressing the terms as decimal quantities, we have S+ = 01 + 001 + 0001 + 00001 + 000001 + The first five terms in the sequence of partial sums S* are S* = 01, 011, 0111, 01111, 011111, 7 We want to find the limit Lim
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if it exists From the results of Problem 6, we see that it s the limit of the sequence of partial sums S* = 01, 011, 0111, 01111, 011111, so we know that Lim
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i =1 n
= 011111
We learned in our algebra courses that 011111 = 1/9 Therefore Lim
1/10i
i =1
= 1/9
576 Worked-Out Solutions to Exercises: 11-19
8 When n = 2, the partial sum is (1 + 2)/22 = 075 When n = 6, the partial sum is approximately (1 + 2 + 3 + 4 + 5 + 6)/62 = 058333 When n = 10, the partial sum is (1 + 2 + 3 + + 8 + 9 + 10)/102 = 055 When n = 20, the partial sum is (1 + 2 + 3 + + 18 + 19 + 20)/202 = 0525 We re approaching 1/2 from the right (the positive side) If you re a computer expert, try programming your machine to work out the partial sums for much larger values of n, and see more clearly that 1/2 (or 05) is indeed the limit of this sequence of partial sums 9 Refer to Table B-1 for squares of integers from 1 to 20 When n = 2, the partial sum is (12 + 22)/23 = 0625
Table B-1 Squares and cubes of positive integers from 1 to 20
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 n2 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 n3 1 8 27 64 125 216 343 512 729 1000 1331 1728 2197 2744 3375 4096 4913 5832 6859 8000
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When n = 6, the partial sum is approximately (12 + 22 + 32 + 42 + 52 + 62)/63 042130 Remember that the wavy equals sign means is approximately equal to When n = 10, the partial sum is (12 + 22 + 32 + + 82 + 92 + 102)/103 = 0385 When n = 20, the partial sum is (12 + 22 + 32 + + 182 + 192 + 202)/203 = 035875 We re approaching 1/3 from the right If you re a computer expert, try programming your machine to work out the partial sums for much larger values of n, and see more clearly that 1/3 (or 033333) is indeed the limit of this sequence of partial sums 10 Refer to Table B-1 for cubes of integers from 1 to 20 When n = 2, the partial sum is (13 + 23)/24 = 05625 When n = 6, the partial sum is approximately (13 + 23 + 33 + 43 + 53 + 63)/64 034028 When n = 10, the partial sum is (13 + 23 + 33 + + 83 + 93 + 103)/104 = 03025 When n = 20, the partial sum is (13 + 23 + 33 + + 183 + 193 + 203)/204 = 027563 We re approaching 1/4 from the right If you re a computer expert, try programming your machine to work out the partial sums for much larger values of n, and see more clearly that 1/4 (or 025) is indeed the limit of this sequence of partial sums
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