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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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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 ?v = v bardblvbardbl = [1, 2,?2] ? 1 + 4 + 4 = bracketleftbigg 1 3 , 2 3 ,? 2 3 bracketrightbigg . Another is ??v = 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

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