K ck x converges if x r so 39 by assumption x k k0 r

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Unformatted text preview: ictly decreasing. Exercise Set 11.2 366 25. (a) Yes: a monotone sequence is increasing or decreasing; if it is increasing, then it is increasing and bounded above, so by Theorem 11.2.3 it converges; if decreasing, then use Theorem 11.2.4. The limit lies in the interval [1, 2]. (b) Such a sequence may converge, in which case, by the argument in Part (a), its limit is ≤ 2. ∞ But convergence may not happen: for example, the sequence {−n}+=1 diverges. n 26. (a) an+1 = |x| |x|n |x| |x|n+1 = = an (n + 1)! n + 1 n! n+1 (b) an+1 /an = |x|/(n + 1) < 1 if n > |x| − 1. (c) From Part (b) the sequence is eventually decreasing, and it is bounded below by 0, so by Theorem 11.2.4 it converges. |x| L = 0. (d) If lim an = L then from Part (a), L = n→+∞ lim (n + 1) n→+∞ |x| = lim an = 0 n→+∞ n! n (e) lim n→+∞ √ √ 2 + 2, 2 + 2 + 2 √ √ √ √ √ (b) a1 = 2 < 2 so a2 = 2 + a1 < 2 + 2 = 2, a3 = 2 + a2 < 2 + 2 = 2, and so on indefinitely. 27. (a) √ 2, (c) a2 +1 − a2 = (2 + an ) − a2 = 2 + an − a2 = (2 − an )(1 + an ) n n n n (d) an > 0 and, from Part (b), an < 2 so 2 − an > 0 and 1 + an > 0 thus, from Part (c), a2 +1 − a2 > 0, an+1 − an > 0, an+1 > an ; {an } is a strictly increasing sequence. n n (e) The sequence is increasing and has 2 as an upper bound so it must converge to a limit L, √ √ 2 + an , L = 2 + L, L2 − L − 2 = 0, (L − 2)(L + 1) = 0 lim an+1 = lim n→+∞ n→+∞ thus lim an = 2. n→+∞ √ 28. (a) If f (x) = 1 (x + 3/x), then f (x) = (x2 − 3)/(2x2 ) and f (x) = 0 for x = 3; the minimum 2 √ √ √ √ value of f (x) for x > 0 is f ( 3) = 3. Thus f (x) ≥ 3 for x > 0 and hence an ≥ 3 for n ≥ 2. √ (b) an+1 − an = (3 − a2 )/(2an ) ≤ 0 for n ≥ 2 since an ≥ 3 for n ≥ 2; {an } is eventually n decreasing. √ 3 is a lower bound for an so {an } converges; lim an+1 = lim 1 (an + 3/an ), (c) n→+∞ n→+∞ 2 √ L = 1 (L + 3/L), L2 − 3 = 0, L = 3. 2 29. (a) The altitudes of the rectangles are ln k for k = 2 to n, and their bases all have length 1 so the sum of their areas is ln 2 + ln 3 + · · · + ln n = ln(2 · 3 · · · n) = ln n!. The area under the n n+1 ln x dx, and curve y = ln x for x in the interval [1, n] is 1 n the interval [1, n + 1] so, from the figure, ln x dx = (x ln x − x) 1 n+1 = n ln n − n + 1 and 1 ln x dx. 1 n n n+1 ln x dx < ln n! < 1 (b) ln x dx is the area for x in 1 ln x dx = (n + 1) ln(n + 1) − n so from 1 Part (a), n ln n − n + 1 < ln n! < (n + 1) ln(n + 1) − n, en ln n−n+1 < n! < e(n+1) ln(n+1)−n , nn (n + 1)n+1 en ln n e1−n < n! < e(n+1) ln(n+1) e−n , n−1 < n! < e en 367 Chapter 11 n 30. n! > √ n 1/n (n + 1)n+1 , en √ n √ 1 (1 + 1/n)(n + 1)1/n (n + 1)1+1/n n! n , 1−1/n < < , < n! < e n e e e1−1/n but 1/n nn en−1 (c) From part (b), 1 → e1−1/n < n! < √ n (1 + 1/n)(n + 1)1/n 1 1 1 n! and → as n → +∞ (why?), so lim =. n→+∞ n e e e e √ nn √ n n n n , n! > 1−1/n , lim 1−1/n = +∞ so lim n! = +∞. n→+∞ e n→+∞ en−1 e EXERCISE SET 11.3 1. (a) s1 = 2, s2 = 12/5, s3 = lim sn = n→+∞ 55 62 312 2 − 2(1/5)n , s4 = sn = = − (1/5)n , 25 125 1 − 1/5 22 5 , converges 2 11 1 3 7 15 (1/4) − (1/4)2n , s2 = , s3 = , s4 = sn = = − + (2n ), 4 4 4 4 1−2 44 lim sn = +∞, diverges (b) s1 = n→+∞ (c) 1 1 1 1 1 3 1 = − , s1 = , s2 = , s3 = , s4 = ; (k + 1)(k + 2) k+1 k+2 6 4 10 3 1 1 1 , lim sn = , converges sn = − 2 n + 2 n→+∞ 2 2. (a) s1 = 1/4, s2 = 5/16, s3 = 21/64, s4 = 85/256 n−1 = 1 1 1 − (1/4)n = 4 1 − 1/4 3 (b) s1 = 1, s2 = 5, s3 = 21, s4 = 85; sn = 4n − 1 , diverges 3 sn = 1 4 1+ 1 + ··· + 4 1 4 1− (c) s1 = 1/20, s2 = 1/12, s3 = 3/28, s4 = 1/8; n sn = k=1 1 1 − k+3 k+4 = 3. geometric, a = 1, r = −3/4, sum = 1 1 − , lim sn = 1/4 4 n + 4 n→+∞ 1 = 4/7 1 − (−3/4) 4. geometric, a = (2/3)3 , r = 2/3, sum = 5. geometric, a = 7, r = −1/6, sum = (2/3)3 = 8/9 1 − 2/3 7 =6 1 + 1/6 6. geometric, r = −3/2, diverges n 7. sn = k=1 1 1 − k+2 k+3 = 1 1 − , lim sn = 1/3 3 n + 3 n→+∞ 1 4 n ; lim sn = n→+∞ 1 3 Exercise Set 11.3 n 8. sn = k=1 n 9. sn = k=1 n+1 10. sn = k=2 = ∞ 11. k=3 1 2 368 1 1 − k+1 2k 2 = 1 1 − , lim sn = 1/2 2 2n+1 n→+∞ 1/3 1/3 − 3k − 1 3k + 2 = 1/2 1 1/2 − = k−1 k+1 2 n+1 k=2 1 = k−2 1 − k−1 n+3 k=4 1 1/3 − , lim sn = 1/6 6 3n + 2 n→+∞ n+1 k=2 1 − k−1 ∞ k=1 4k+2 = 7k−1 k=2 1 k+1 1 1 1 1 1 = 1+ − − ; k−1 2 2 n+1 n+2 lim sn = n→+∞ 3 4 ∞ 1/k , the harmonic series, so the series diverges. k=1 12. geometric, a = (e/π )4 , r = e/π < 1, sum = 13. n+1 ∞ 64 k=1 4 7 e4 (e/π )4 =3 1 − e/π π (π − e) k−1 ; geometric, a = 64, r = 4/7, sum = 64 = 448/3 1 − 4/7 14. geometric, a = 125, r = 125/7, diverges 15. 0.4444 · · · = 0.4 + 0.04 + 0.004 + · · · = 0.4 = 4/9 1 − 0.1 16. 0.9999 · · · = 0.9 + 0.09 + 0.009 + · · · = 0.9 =1 1 − 0.1 17. 5.373737 · · · = 5 + 0.37 + 0.0037 + 0.000037 + ·...
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