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Stewart6ET11-2solnChallenging

# Stewart6ET11-2solnChallenging - ’41 Wm M twwt 45)“...

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Unformatted text preview: ’41-; Wm M twwt 45)“) Q9103 )70) 73a. 65. The seriesl — 1+1 — 1+1 — 1+--- diverges (geometric series withr = —-1) so we cannot say that 0=1—l+1~1+l-l+---. 66. If 2 an is convergent, then lim an: Oby Theorem6, so lim — 175 0, and so 21 — is divergent by the Test for ”=1 A "“‘°° n—mo an n=1an Divergence. W ' - u 67‘ 271:1 Ca" 2 1111: 22:1 ca,- = hm CELT-L1 ai = C hm 2,7; 111: = 622:1 an, which exists by hypothesis. 68. If E can were convergent, then 2(1/0) (can) = 2 an would be also, by Theorem 8. But this is not the case, so 2 can must diverge. ‘ 69. Suppose on the contrary that 201,, + b”) converges. Then 201.7, + bn) and 2 an are convergent series. So by Theorem 8, l\ 2 [(an + bn) * an] W0111d 3130 be convergent. But 2 [(011 + bu) — an] = 2 bn, a contradiction, since 2 b, is given to be divergent. \ . . ‘ ‘1‘ 70. No. For example, take 2 an 2 E n and E b = Z(—-n), which both diverge, yet 201,, + bn) = E 0, which converges ‘ with sum 0. ' ' 1 71. The partial sums {31,} form an increasing sequence, since 8" — sn 1— ~ an > 0 for all n. Also, the sequence {sn}1s bounded \ since 8,, < 1000 for all 77,. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series 2 an is convergent 73. (a) At the ﬁrst step, only the interval (— 3, 3) (length— 35') is removed At the second step, we remove the intervals (9, 9) and ('5’, g), which have a total length of 2 . (5)2 . At the third step, we remove 22 intervals, each of length (if? . In general, at the nth step we remove 2"“1 intervals, each oflength (é) 11, for a length of?“1 - (-31)" = §(§)"_1. Thus, the total W length of all removed intervals is E- gg )" 1:1 g 1/323 = l [geometric series With a— — g and r = g]. Notice that at n=1 the nth step, the leﬂmost interval that is removed is (ﬁ) " , (g) n) , so we never remove O, and 0 is in the Cantor set. Also, the rightmost interval removed is (1 ~ (g) n , 1 — (ﬁ) n), so 1 is never removed. Some other numbers in the Cantor set 1 2 1 2 7 s arcs,3,9,9,-§,and9. (b) The area removed at the ﬁrst step is— 9; at the second step,8 -(-1-)2' , at the third step, (8)2- -%( )3. In general, the area removed at the nth step is (8)“ 1()n= 91% )n1,so the total area of all removed squares is i 5(3)” 1 = 1 i/:/9 2 1' n=1 ...
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