I}\ 11 J Qo = 250. D2 = ho + 2hl = (~) = 25 feet.

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1 J Qo = t (time, days) 23. (a) Let hn be the height of the nth bounce after the ball hits the floor for the nth time. Then from Figure 9., ho = height before first bounce = 0 feet, hl = height after first bounce = 0 ( ~ ) feet, h2 = height after second bounce = 0 (~)2feet. Generalizing gives I}\ Figure 9. (b) When the ball hits the floor for the first time, the total distance it has traveled is just Dl = 0 feet. (Notice that this is the same as ho = 0.) Then the ball bounces back to a height of hl = 0 (~), comes down and hits the floor for the second time. See Figure 9.. The total distance it has traveled is D2 = ho + 2hl = (~) = 25 feet. Then the ball bounces back to a height of h2 = 0 ( ~ ) 2, comes down and hits the floor for the third time. It has traveled

2 546 Chapter Nine /SOLUTIONS Similarly, D4 = ho + 2h + 2h2 + 2hs = (i) (i) (i) s = (i) s ~ feet. (c) When the ball hits the floor for the nth time, its last bounce was of height hn-l. (b). we get Dn = ho + 2h + 2h2 + 2h hn-l = 0 + _2.0 (~) (~) (~) 3 + Thus, by the method used in part = (~) - (~) = ( -4 (3)n-l 24. (a) The acceleration of gravity is 32 ftlsec2 so acceleration = 32 and velocity v = 32t + C. Since the ball is dropped, its initial velocity is 0 so v = 32t. Thus the position is s = 6t2 + C. Calling the initial position s = 0, we have s = 6t. The distance traveled is h so h = 6t. Solving for t we get t = ivh. (b) The first drop from 0 feet takes i.f[{j seconds. The first full bounce (to 0. (~) feet) takes i~ seconds to rise, therefore the same time to come down. Thus, the full bounce, up and down, takes 2(i>~ seconds. The next full bounce takes 2(i>0.(~>2 = 2( i >.f[{j (.Jf) 2 seconds. The nth bounce takes 2(i>.f[{j I.Jf) n seconds. Therefore the Total amount of time./i) 3 + Geometric series with a = t.;w.jf =!.;w.jf and x =.Jf 25. (a) Total amount of money deposited = = + (0.92) + (0.92)2 + = -- = 250 dollars (b) Credit multiplier = 250/lOO = 2.50 The 2.50 is the factor by which the bank has increased its deposits, from $ to $ The amount of additional income generated directly by people spending their extra money is $(0.8) = $80 million. This additional money in turn is spent, generating another ($(0.8)) (0.8) = $(0.8)2 IiIillion. This continues indefinitely, resulting in Total additional income = (0.8) + (0.8t + (0.8)3 +. ~ = $4 million = -0.8

3 Solutions for Section 9.2 Exercises I. Since lim xn = if Ixl < and 0.2 <, we have lim (0.2)n = 0. n-+ n-+ 2. Since 2n increases without bound as n increases, the limit does not exist. 3. Since lim xn = if Ixl < and I -0.3 <, we have lim ( -0.3)n = 0. n-+ n-+ 4. Since tim xn = if Ixl < and le-2 <, we have lim (e-2n) = tim (e-2)n = 0, so lim (3+e-2n) = 3+0 = 3. n-+oo n-+ n-+ n-+ 5. Since Sn = cog( 7rn ) = if n is even and Sn = cos( 7rn ) = - if n is odd, the values of Sn oscillate between and -, so the limit does not exist. 6. Since lim xn = iflxl < and -32 <, we have lim (~ ) = lim ( ~ ) n = 0. n-+oo n-+oo 3n n-+oo 3 7. As n increases, the term 4n is much larger than 3 and 7n is much larger than 5. Thus dividing the numerator and denominator by n and using the fact that lim l/n = 0, we have n-+ I " 3 + 4n Imn-+CX) 5 + 7n = Jim (3/n) +4 = ~..-+ (5/n) As n increases, the tend 2n is much larger in magnitude than ( -)n5 and the tend 4n is much larger in magnitude than ( -)n30 Thus dividing the numerator and denominator by n and using the fact that lim l/n = 0. we have n-+cx> lim 2n+(-)n5 = lim 2+ (-)n5/n =.!0 n-+cx>4n-{-)n3 n-+cx>4-(-)n3/n 2 9. We use the integral test to determine whether this series converges or diverges. We determine whether the corresponding / ~dx converges or diverges: improper integral x l O. l b.-l i b. ( - ) 3dx = hm 3dx = hm -2 2 = lim -2b = - 2. X b-+-oo X b-+-oo X b-+-oo / Since the integral ;SdX converges, we conclude from the integral test that the series L ~ converges. n=l 0. We use the integral test to determine whether this series converges or diverges. We determine whether the corresponding improper integral / -;- x + dx converges or diverges: Since the integral "'

4 9. Let an = (lnn)/nand f(x) = (lnx)/x.we use the integratest and consider the improperintegra l oo ~dx. 9.2 SOLUTIONS 549 c x and In R grows without bound as R -t, the integral diverges. Therefore, the integral test tells us that the series, ~ Inn al di L -;:;-' so verges. n=l 20. We use the integral test and calculate the corresponding improper integral,!3(x + )/(x2 + 2x + 2) dx: «) Since the limit does not exist (it is ), the integral diverges, so the series ~ 2 n +2 L..,n + n+ 2 diverges. n=3 2. Using left-hand sums for the integral of f(x) = /(4::- 3) over the interval 5 ::5 n + with uniform subdivisions of length I gives a lower bound on the partial sum: 8" = n-3 > I n+l -= -In(4x -3) "+ = -In(4n+) dx 5 9 4x Since In( 4n + ) increases without bound as n -+, the partial sums of the series are unbounded. Thus, this is not a convergent series. 22. Using right-hand sums for the integral of f(x) = X-3/2 over the interval $: x $: n with uniform subdivisions of length I gives: --! " -3/ < :I: d::=- 2( n -/2 23/2 n3/2 -). Adding to both sides gives an upper bound on the partial sum 8n=I <-2n ( -/2 - ) 23/2 n3/2 Thus, as n -t, the sequence of partial sums is bounded. Each successive partial sum is obtained from the previous one by adding one more term in the series. Since all the terms are ~sitive, the sequence of partial sums is increasing. Hence the series converges. roo 23. (a) We compare L l/np with the integral (/::P)d::. For p #, we have n= / i b -p+l I b b-p+l - -dx = lim -dx = tim =-- = tim. xp b-+oo xp b-+oo -p + b-+oo -p + If p >, the power of b is negative, so this limit exists. Thus the integral converges, so the series converges. (b) If p <, then the power of b is positive and the limit does not exist. Thus, the integral diverges, so the series diverges. We have to look at the case p = separately, since the fond of the antiderivative is different in that case. If p =, we compare L l/n with / (l/x)dx. Since n=l /!dx = lim / b!dx = lim Inlxl lb = lim Inb, X b-+oo X b-+oo b-+oo and since lim In b does not exist, the integral diverges, so the series diverges. Combining these results shows that b-+oo L l/np diverges if p $. n=l

5 556 Chapter Nine /SOLUTIONS ~ Therefore ) ~ = r;; + )en+ = nen = (-.!!.--!. /ani.-.!.- (n+l)en+l n+l e nen L=lim~_I n-+cxi /an/ -; < I. Since L <, the ratio test tells us that ~ 2- converges. L., ne" "= 6. Let a" = /(2n + ). Then replacing n by n + gives a,,+l = /(2n + 3). Since 2n + 3 > 2n +, we have O < a,,+l = -.< ~ = an, 2n + 3 2n + Problems 9. The partial sums are 8 =, 82 = -, 83 = 2, 80 = -5, 8 = 6, 8 = -50, 80 = 5, 80 = -5, 8 = 50, which appear to be oscillating further and further from 0. This series does not converge. 20. The partial sums look like: 8 =, 82 = 0, 83 = 0.5, 84 = , 85 = 0.375, 80 = , 820 = , and higher partial sums agree with these first 4 decimal places. The series appears to be converging to about Since an = l/n! is positive and decreasing and limn-+~ l/n! = 0, the alternating series test confirms the convergence of this series. 2. The first few terms of the series may be written e +e +e +...; this is a geometric series with a = and x = e-l = l/e. Since /x/ <, the geometric series converges to e 8 - -x - -e-l -e We use the ratio test and calculate Im-- -(0_)n+l Im I(n + )! " In " /an+l - Since the limit is less than, the series converges- 23. We use the ratio test and calculate n-+oo lanl n-+ (0_)n In! n-+oo n + lim ~ = lim ~(n +Z = lim (~ --L ) = lim ( n-+oo lanl n-+oo (n- )!ln2 n-+oo (n- )! (n + )2 n-+ Since the limit does not exist (it is ), the series diverges- 24. The first few terms of the series may be written n. n2 (n + ) e+e +e +.--=e+e-e+e-e +---; this is a geometric series with a = e and x = e. Since /xi >, this geometric series diverges"

6 9.3 SOLUTIONS Let an = /.;3n-=-T. Then replacing n by n + gives an+l = l/y3(n+ ) -. Since y3(n+l)-i>~, we have an+l < an. In addition, lim..-+~ a.. = 0 so the alternating series test tells us that the series ~ ~ converges. L v 3n -..= 26. Since the exponential, 2.., grows faster than the power, n2, the terms are growing in size. Thus, lim a.. # 0. We conclude..-+~ that this series diverges. 27. Let a.. = n(n + )/vn3 + 2n2. Since n3 + 2n2 = n2(n + 2), we have am = n(n+) - n+ -,.n.;n+2 -=.;n+2 - SO an grows without bound as n -+, therefore the series L ~ diverges. yn- + 2n2 n=l 28. Let an = / vn2(n + 2). Since n2(n + 2) = n3 + 2n2 > n3, we have O < an < ;;:372". aso converges. 29. (a) Assume that n is even. Then Lp(n+"i) n=l -! +! -! +! = (-! ) + (! -! ) (--.!.-- -! n n- n = (n-).n' ). -.!.-+ (b) The given senes N + H is term by term less than the series = Since this second series, L l/n2, converges by the integral test, the first series converges. (c) By parts (a) and (b), the sequence of partial sums for even n converges. The partial sum for odd n equals l/n plus the partial sum for even n -. Thus the partial sums for odd n approach the partial sums for even n, as n -.. Therefore the sequence of all partial sums converges, and hence the series converges. 30. The argument is false. Property of Theorem 9.2 only applies to convergent series. Furthermore, since n(n + ) > n2 we can compare L ~ with the convergent series L ~ and deduce that it converges. n= n= 3. Suppose we let Cn = (-I)nan. (We have just given the terms of the series L(-I)nan a new name.) Then ICnI = (-I)nanl = lanl. Thus E ICn I converges, and by Theorem 9.5, LCn = L<-l)nan converges.