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abdallah (haa2348) – Assignment 9 Due by 11:59pm on Sun 4/26 – reyes – (52535) 1This print-out should have 35 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.Taylor series, power series, some three di-mensional graphing001 10.0 pointsFind the interval of convergence of thepower series∞Xn = 12xn√n.1. [−2, 2)2. (−2, 2]3. (−1, 1)4. [−2, , 2]5. [−1, 1]6. (−1, 1]7. [−1, 1) correct8. (−2, 2)Explanation:Whenan=2xn√n,thenan+1an=xn+1√n + 1·√nxn=|x|√n√n + 1= |x|rnn + 1.Thuslimn → ∞an+1an= |x|.By the Ratio Test, therefore, the given seriesconverges when |x| < 1, and diverges when|x| > 1.We have sti ll to check for convergence atx = ±1. But w hen x = 1 , the series reducesto∞Xn = 12√nwhich diverges by the p-series test with p =12≤ 1. On the other hand, when x = −1, theseries reduces to∞Xn = 1(−1)n2√nwhich converges by the Alternating SeriesTest. Thus theinterval of convergence = [−1, 1) .002 10.0 pointsDetermine the radius of convergence, R, ofthe power series∞Xn = 12xn√n.1. R = 02. R = 1 correct3. R =124. R = 25. R = ∞Explanation:The given series has the form∞Xn = 1anxnwithan=2√n.Now for this series,abdallah (haa2348) – Assignment 9 Due by 1 1 :59pm on Sun 4/26 – reyes – (52535 ) 2(i) R = 0 if it converges only at x = 0,(ii) R = ∞ if it converges for al l x,while 0 < R < ∞(iii) if it converges when |x| < R, and(iv) diverges when |x| > R.Butlimn→∞an+1an= limn→∞√n√n + 1= 1 .By the Ratio Test, therefore, t he g iven series(a) converges for all |x| < 1, and(b) diverges for all |x| > 1.Consequently, the given series has radius ofconvergenceR = 1 .003 10.0 pointsFind the interval of convergence of the se-ries∞Xn = 1xn√n + 3.1. int erval of cgce = [−1, 1]2. int erval of cgce = [−1, 1) correct3. int erval of cgce = [−3, 3]4. int erval of cgce = (−1, 1)5. converges onl y a t x = 06. int erval of cgce = (−3, 3]Explanation:Whenan=xn√n + 3,thenan+1an=xn+1√n + 4√n + 3xn= |x|√n + 3√n + 4= |x|rn + 3n + 4.Butlimn → ∞n + 3n + 4= 1,solimn → ∞an+1an= |x|.By the Ratio Test, therefore, the given series(i) converges when |x| < 1,(ii) diverges when |x| > 1.We have still to check what happens at theendpoints x = ±1. At x = 1 the seriesbecomes(∗)∞Xn = 11√n + 3.Applying the Integral Test withf(x) =1√x + 3we see that f is continuous, positive, and de-creasing on [1, ∞), but the improper integralI =Z∞1f(x) dxdiverges, so t he infinite series (∗) divergesalso.On the other hand, at x = −1, the seriesbecomes(‡)∞Xn = 1(−1)n√n + 3.which is an alternating series∞Xn = 1(−1)nan, an= f(n)withf(x) =1√x + 3abdallah (haa2348) – Assignment 9 Due by 1 1 :59pm on Sun 4/26 – reyes – (52535 ) 3the same continuous, positive and decreasingfunction a s before. Sincelimx → ∞f(x) = limx → ∞1√x + 3= 0 ,however, the Alternating Series Test ensuresthat (‡) converges.Consequently, theinterval of convergence = [−1, 1) .004 10.0 pointsDetermine the interval of convergence ofthe series∞Xn = 1(−1)n4n(n + 6)xn.1. (−∞, ∞)2. none of the other answers3. [−6, 6)4. converges only at x = 05.−16,166.−14,14i7. (−4, 4] correctExplanation:The given series has the form∞Xn = 1anxnwithan=(−1)n4n(n + 6).Now for this series,(i) R = 0 if it converges only at x = 0,(ii) R = ∞ if it converges for all x,while if R > 0,(iii) it converges when |x| < R, and(iv) diverges when |x| > R.Butlimn → ∞an+1an= limn → ∞n + 64(n + 7)=14.By the Ratio Test, therefore, the given seriesconverges when |x| < 4 and diverges when|x| > 4. On the other hand, at the pointsx = −4 and x = 4 the series reduces to∞Xn = 11n + 6,∞Xn = 1(−1)nn + 6respectively. Now, by the p-series Test withp = 1, the first of these series diverges, whileby the Alternating Series Test, the secondconveges. Consequently, the given series hasinterval of convergence = (−4, 4] .keywords:005 10.0 pointsFind the interval of convergence of the se-ries∞Xn = 1(−1)nxn4n + 5.1. interval of cgce = [−5, 4]2. interval of cgce = (−4, 5]3. converges only at x = 04. interval of cgce = [−1, 1)5. interval of cgce = [−1, 1]6. interval of cgce = (−1, 1] correct7. interval of cgce = (−1, 1)abdallah (haa2348) – Assignment 9 Due by 1 1 :59pm on Sun 4/26 – reyes – (52535 ) 48. int erval of cgce = (−∞, ∞)Explanation:Whenan= (−1)nxn4n + 5,thenan+1an=−xn+14n + 94n + 5xn= |x|4n + 54n + 9.Butlimn → ∞4n + 54n + 9= 1,in which caselimn → ∞an+1an= |x|.By the Ratio Test, therefore, t he g iven series(i) converges when |x| < 1,(ii) diverges when |x| > 1.We have still to check what happens at theendpoints x = ±1. At x = −1 the seriesbecomes∞Xn = 114n + 5which diverges by the Inegral Test. On theother hand, at x = 1, the series becomes∞Xn = 1(−1)n4n + 5which converges by the Alternating SeriesTest.Consequently, theinterval of convergence = (−1, 1] .006 10.0 pointsFind the interval of convergence of the se-ries∞Xn = 1(−1)nxn2n2+ 1.1. [−1, 1)2. [−2, 1]3. (−1, 1]4. converges only at x = 05. (−2, 1]6. [−1, 1] correctExplanation:Whenan= (−1)nxn2n2+ 1,thenan+1an=−xn+12(n + 1)2+ 12n2+ 1xn= |x|2n2+ 12(n + 1)2+ 1.But2(n + 1)2+ 1 = 2n2+ 4n + 3 ,whilelimn → ∞2n2+ 12n2+ 4n + 3= 1 .Thuslimn → ∞an+1an= |x|.By the Ratio Test, therefore, the given series(i) converges when |x| < 1,(ii) diverges when |x| > 1.We have still to check what happens at theendpoints x = ±1. At x = −1 the seriesbecomes(∗)∞Xn = 112n2+ 1.Applying the Integral Test withf(x) =12x2+ 1abdallah (haa2348) – Assignment 9 Due by 1 1 :59pm on Sun 4/26 – reyes – (52535 ) 5we see that f is continuous, positive, and de-creasing on [1, ∞); in addition,the improperintegralI =Z∞1f(x) dxconverges, so the infinite series (∗) convergesalso.On the other hand, at x = 1, the seriesbecomes(‡)∞Xn = 1(−1)n2n2+ 1.which is an alternating series∞Xn = 1(−1)nan, an= f(x)withf(x) =12x2+ 1the same continuous, positive and decreasingfunction a s before. Aslimx → ∞f(x) = limx → ∞12x + 1= 0

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