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Exam 3 Solutions (form A) Spring 2006.pdf

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- Texas
- Texas A&M University
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- Mathematics 152
- Albrecht/austin
- Exam 3 Solutions (form A) Spring 2006.pdf

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Labs with Maple for Single Variable Calculus Concepts and Contexts, 3rd edi...

Spring 2006 Math 152 Exam 3A: Solutions Mon, 01/May c©2006, Art Belmonte Notes • The notation b n ↓ 0 means 0 < b n+1 ≤ b n and lim b n = 0; i.e., the sequence {b n } decreases to zero in the limit. • GFF [“Generalized Fun Fact”]: Let p (n) = m summationdisplay k=0 c k n k be a polynomial in n with c m > 0. Then lim n→∞ n radicalbig p (n) = 1. This boosts the power of the Root Test, which was optionally covered in some classes. (In most classes, only the Ratio Test was covered.) 1. (b) Algebraic manipulation gives lim n→∞ parenleftBigg n 2 + 1 n 2 sin parenleftbigg pin 2n + 1 parenrightbigg parenrightBigg = lim n→∞ parenleftBigg 1 + 1 n 2 1 sin parenleftBigg pi 2 + 1 n parenrightBiggparenrightBigg = sin parenleftBig pi 2 parenrightBig = 1. 2. (d) Only the series summationtext (−1) n 4 √ n converges, but not absolutely. • I. The series summationtext (−1) n n 8 converges absolutely since summationtext |a n | = summationtext 1 n 8 , a convergent p-series (p = 8 > 1). • II. The series summationtext (−1) n+1 parenleftBig n 5 +2 n 3 parenrightBig diverges by the Test for Divergence since lim a n negationslash= 0. (Indeed, the limit does not exist since lim sup a n = ∞ and lim inf a n = −∞.) • III. The alternating series summationtext (−1) n 4 √ n converges by the Alternating Series Test since b n = |a n | = 1 4 √ n ↓ 0. Note, however, that summationtext |a n | = summationtext 1 n 1/4 is a divergent p-series (p = 1 4 ≤ 1). Accordingly, summationtext (−1) n 4 √ n converges, but is not absolutely convergent. (We say it is conditionally convergent.) 3. (b) We have 1 1− parenleftbig −x 4 parenrightbig = ∞ summationdisplay n=0 parenleftBig −x 4 parenrightBig n = ∞ summationdisplay n=0 (−1) n x 4n , provided vextendsingle vextendsingle −x 4 vextendsingle vextendsingle < 1 or |x| < 1. 4. (d) The series ∞ summationdisplay n=1 1 n (n + 1) is a telescoping sum. • First do a partial fraction decomposition. 1 n (n + 1) = A n + B n + 1 1 = A (n + 1) + Bn 0n + 1 = (A + B) n + A Thus A + B = 0 and A = 1, whence B = −A = −1. Therefore, ∞ summationdisplay n=1 1 n (n + 1) = ∞ summationdisplay n=1 parenleftbigg 1 n − 1 n + 1 parenrightbigg . • Now look at the sequence of partial sums to determine its limit s, the sum of the series. s 1 = 1− 1 2 s 2 = parenleftBig 1− 1 2 parenrightBig + parenleftBig 1 2 − 1 3 parenrightBig = 1− 1 3 . . . s n = 1− 1 n+1 s = lim n→∞ s n = 1 5. (c) Only statement III is true. • I. If lim b n = 0, then summationtext b n converges. This is false. A counterexample is the harmonic series summationtext 1 n , a divergent p-series (p = 1 ≤ 1). • II. If 0 ≤ a n ≤ b n and summationtext b n diverges, then summationtext a n diverges. This is false. A countexample is summationtext a n = 1 n 2 and summationtext b n = summationtext 1 n . Note that summationtext 1 n , the harmonic series, diverges. Yet while 0 ≤ a n = 1 n 2 ≤ 1 n = b n , we see that summationtext 1 n 2 converges (p-series with p = 2 > 1). • III. If summationtext a n converges, then lim a n = 0. This is true. As stated in Section 10.2, this condition is necessary for a series to converge. 6. (d) Compute the sum of this geometric series via the Geometric Series Theorem. ∞ summationdisplay n=1 2 n 3 n = ∞ summationdisplay n=1 parenleftbigg 2 3 parenrightbigg n = ∞ summationdisplay n=1 parenleftbigg 2 3 parenrightbiggparenleftbigg 2 3 parenrightbigg n−1 = a 1−r = 2/3 1− 2 3 = 2/3 1/3 = 2 7. (d) The series ∞ summationdisplay n=3 1 n ln n diverges by the Integral Test. integraldisplay ∞ 3 1 x ln x dx = lim t→∞ integraldisplay t 3 1 ln x 1 x dx = lim t→∞ integraldisplay ln t ln 3 1 u du (sub u = ln x) = lim t→∞ ln u vextendsingle vextendsingle vextendsingle ln t ln 3 = lim t→∞ (ln (ln t)−ln (ln 3)) = ∞ 8. (d) Note that − 1 n ≤ sin n n ≤ 1 n . Since lim n→∞ parenleftbigg − 1 n parenrightbigg = 0 and lim n→∞ 1 n = 0, we conclude that lim n→∞ sin n n = 0 by the Squeeze Theorem. 9. (c) The series ∞ summationdisplay n=1 x 2n √ n converges for −1 < x < 1 as follows. 1 • The Ratio Test gives lim n→∞ vextendsingle vextendsingle vextendsingle vextendsingle a n+1 a n vextendsingle vextendsingle vextendsingle vextendsingle = lim n→∞ parenleftBigg |x| 2n+2 √ n + 1 √ n |x| 2n parenrightBigg = lim n→∞ radicalBigg 1 1 + 1 n |x| 2 = |x| 2 need < 1 =⇒ |x| < 1. Hence the radius of convergence is R = 1. • [Alternatively, use the Root Test along with the GFF. As n →∞, we have n radicalbig |a n | = parenleftbig |x| 2n parenrightbig 1/n n radicalbig √ n = |x| 2 radicalbig n √ n → |x| 2 need < 1, whence |x| < 1 and R = 1.] • At the endpoints x = ±1, the series is summationtext 1 n 1/2 , a divergent p-series (p = 1 2 ≤ 1). So the interval of convergence is I = (−1, 1). 10. (a) The series ∞ summationdisplay n=1 a n = ∞ summationdisplay n=1 1 (1 + n) n p = ∞ summationdisplay n=1 1 n p + n p+1 converges for p > 0 by the Limit Comparison Test. • Let summationdisplay b n = summationdisplay 1 n p+1 . Then lim n→∞ a n b n = lim n→∞ n p+1 (1 + n) n p = lim n→∞ 1 1 + 1 n = 1 > 0. Since summationdisplay 1 n p+1 converges only if q = p + 1 > 1 or p > 0, we conclude by the Limit Comparison Test that summationdisplay 1 (1 + n) n p converges only if p > 0. [Brian Winn] • Here is an alternative proof that employs repeated use of the Comparison Test instead. – First, ∞ summationdisplay n=1 1 n p + n p+1 ≤ ∞ summationdisplay n=1 1 n p+1 = ∞ summationdisplay n=1 1 n q , which converges provided p + 1 = q > 1 or p > 0. Therefore, ∞ summationdisplay n=1 1 (1 + n) n p converges for p > 0 by the Comparison Test. – Next, ∞ summationdisplay n=1 1 n p + n p+1 ≥ ∞ summationdisplay n=1 1 n p+1 + n p+1 = 1 2 ∞ summationdisplay n=1 1 n p+1 = 1 2 ∞ summationdisplay n=1 1 n q , which diverges for p + 1 = q ≤ 1 or p ≤ 0. Therefore, ∞ summationdisplay n=1 1 (1 + n) n p diverges for p ≤ 0 by the Comparison Test. – Accordingly, ∞ summationdisplay n=1 1 (1 + n) n p converges only when p > 0. 11. The power series ∞ summationdisplay n=1 (x −3) n n 4 n has radius of convergence R = 4 and interval of convergence I = [−1, 7). • The Ratio Test gives lim n→∞ vextendsingle vextendsingle vextendsingle vextendsingle a n+1 a n vextendsingle vextendsingle vextendsingle vextendsingle = lim n→∞ parenleftBigg |x −3| n+1 (n + 1) 4 n+1 n4 n |x −3| n parenrightBigg = lim n→∞ parenleftBigg 1 1 + 1 n parenrightBigg |x −3| 4 = |x −3| 4 need < 1 =⇒ |x −3| < 4. Hence the radius of convergence is R = 4. • [Alternatively, use the Root Test along with the GFF. As n →∞, we have n radicalbig |a n | = |x −3| 4 n √ n = |x −3| 4 need < 1, whence |x −3| < 4 and R = 4.] • At the left endpoint x = −1, we have the alternating series summationtext (−1) n n , which converges by the Alternating Series Test since b n = |a n | = 1 n ↓ 0. • At the right endpoint x = 7, we have the harmonic series summationtext 1 n , a divergent p-series (p = 1 ≤ 1). • So the interval of convergence is I = [−1, 7). 12. (a) The series ∞ summationdisplay n=1 e n n! converges absolutely (and hence converges) by the Ratio Test. lim n→∞ vextendsingle vextendsingle vextendsingle vextendsingle a n+1 a n vextendsingle vextendsingle vextendsingle vextendsingle = lim n→∞ parenleftBigg e n+1 (n + 1)! · n! e n parenrightBigg = lim n→∞ e n + 1 = 0 < 1 (b) The alternating series ∞ summationdisplay n=1 (−1) n 1 + ln n converges by the Alternating Series Test since b n = vextendsingle vextendsingle vextendsingle 1 a n vextendsingle vextendsingle vextendsingle = 1 1+ln n ↓ 0. (c) The series ∞ summationdisplay n=1 n n 2.001 + 4 converges by the Comparison Theorem since 0 ≤ n n 2.001 + 4 ≤ n n 2.001 = 1 n 1.001 and summationtext 1 n 1.001 is a convergent p-series (p = 1.001 > 1). 13. Let f (x) = e 3x and a = 2. Then f (n) (x) = 3 n e 3x and f (n) (2) = 3 n e 6 . So the Taylor series for f (x) = e 3x at a = 2 is ∞ summationdisplay n=0 f (n) (2) n! (x −2) n = ∞ summationdisplay n=0 3 n e 6 n! (x −2) n . 2 14. (a) Recall that sin z = ∞ summationdisplay n=0 (−1) n z 2n+1 (2n + 1)! for z ∈R. Thus integraldisplay 1 0 sin parenleftBig x 2 parenrightBig dx = integraldisplay 1 0 ∞ summationdisplay n=0 (−1) n parenleftBig x 2 parenrightBig 2n+1 (2n + 1)! dx = integraldisplay 1 0 ∞ summationdisplay n=0 (−1) n x 4n+2 (2n + 1)! dx = ∞ summationdisplay n=0 (−1) n x 4n+3 (2n + 1)! (4n + 3) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x=1 x=0 = ∞ summationdisplay n=0 (−1) n 1 4n+3 (2n + 1)! (4n + 3) − ∞ summationdisplay n=0 0 = ∞ summationdisplay n=0 (−1) n (2n + 1)! (4n + 3) (b) The Alternating Series Estimation Theorem gives |R 9 | ≤ |a 10 | = 1 10 2 = 1 100 . 15. (a) One unit vector that is parallel to v = [1, 2,−2] is ˆv = v bardblvbardbl = [1, 2,−2] √ 1 + 4 + 4 = bracketleftbigg 1 3 , 2 3 ,− 2 3 bracketrightbigg . Another is −ˆv = bracketleftbigg − 1 3 ,− 2 3 , 2 3 bracketrightbigg . (b) The vector (parallel) projection of b = [4, 2, 0] onto a = [1,−1, 1] is proj a b = parenleftbigg a · b bardblabardbl parenrightbigg a bardblabardbl = parenleftbigg 4−2 + 0 √ 1 + 1 + 1 parenrightbigg [1,−1, 1] √ 3 = 2 3 [1,−1, 1] = bracketleftbigg 2 3 ,− 2 3 , 2 3 bracketrightbigg . 3