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UCLA MATH 31B - Practice Problems for Midterm 2

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Practice Problems for Midterm 2(1) Compute the radius of convergence of the series∞Xn=14nn(x − 3)nSolution. We will determine the radius of convergence by applyingthe ratio test. We havelimn→∞4n+1n+1(x − 3)n+14nn(x − 3)n= limn→∞4(x − 3)1n+11n= limn→∞4(x − 3)n + 1n= 4(x − 3)since limn→∞nn+1= 1. By the ratio test, the seriesP∞n=14nn(x − 3)ndiverges if |4(x − 3)| > 1 and converges when|4(x − 3)| < 1 ⇐⇒ |x − 3| <14,so the radius of convergence is14.(2) Compute the radius of convergence of the series∞Xn=1xn2nn2Solution. We havelimn→∞xn+12n+1(n+1)2xn2nn2= limn→∞x2·1(n+1)21n2= limn→∞x2nn + 12=x212since limn→∞nn+12= 1. By the ratio test, the series diverges if|x2| > 1 and converges whenx2< 1 ⇐⇒ |x| < 2,so the radius of convergence is 2.(3) Compute the radius of convergence of the series∞Xn=12xn3nn2Solution. We havelimn→∞2xn+13n+1(n+1)22xn3nn2= limn→∞x3·1(n+1)21n2= limn→∞x3nn + 12=x3since limn→∞nn+12= 1. By the ratio test, the series diverges if|x3| > 1 and converges whenx3< 1 ⇐⇒ |x| < 3,so the radius of convergence is 3.(4) Calculate the sum∞Xn=1(−1)n2n+13nSolution. This is a constant multiple of a geometric series. Factoringout the 2, we obtain∞Xn=1(−1)n2n+13n= 2∞Xn=1(−1)n2n3n.3Note(−1)n2n3n= (−23)n, so this is equal to2∞Xn=1−23n.We factor out−23and then apply the geometric series formula since|−23| < 1.2−23∞Xn=0−23n= −43·11 − (−23)= −43(1 +23)= −45.(5) Calculate the sum∞Xn=3(−1)n3n+14nSolution. Factoring out the 3, we obtain∞Xn=3(−1)n3n+14n= 3∞Xn=3(−1)n3n4n.Note(−1)n3n4n= (−34)n, so this is equal to3∞Xn=3−34n.4We factor out−343= −2764and then apply the geometric seriesformula since |−34| < 1.3−2764∞Xn=0−34n= −8164·11 − (−34)= −8164(1 +34)= −81112.(6) Calculate the sum∞Xn=02n+ (−1)n3nSolution. First we distribute the sum to obtain∞Xn=02n+ (−1)n3n=∞Xn=02n3n+∞Xn=0(−1)n3n=∞Xn=023n+∞Xn=0−13n.Since |23| < 1 and |−13| < 1, the geometric series formula tells us thatthe above is equal to11 −23+11 − (−13)=113+143= 3 +34=154.(7) Calculate the sum∞Xn=35nn + 3−5(n + 1)n + 45Solution. This is a telescoping series. For M ≥ 3 letSM=MXn=35nn + 3−5(n + 1)n + 4= 5MXn=3nn + 3−n + 1n + 4denote the M-th partial sum. The first four partial sums areS3= 536−47S4= 536−47+47−58S5= 536−47+47−58+58−69S6= 536−47+47−58+58−69+69−710Notice all terms cancel except the first positive term36that comesfrom plugging in n = 3 and the last negative term −M+1M+4that comesfrom plugging in n = M. ThereforeSM= 536−M + 1M + 4.Since limM→∞M+1M+4= 1, we have∞Xn=35nn + 3−5(n + 1)n + 4= limM→∞SM= 536− 1= −52.(8) Calculate the sum∞Xn=1ln(n + 1)n + 2−ln(n + 2)n + 3Solution. This is a telescoping series. For M ≥ 3 letSM=MXn=1ln(n + 1)n + 2−ln(n + 2)n + 36denote the M-th partial sum. The first four partial sums areS1=ln 23−ln 34S2=ln 23−ln 34+ln 34−ln 45S3=ln 23−ln 34+ln 34−ln 45+ln 45−ln 56S4=ln 23−ln 34+ln 34−ln 45+ln 45−ln 56+ln 56−ln 67Notice all terms cancel except the first positive termln 23that comesfrom plugging in n = 3 and the last negative term −ln(M+2)M+3thatcomes from plugging in n = M. ThereforeSM=ln 23−ln(M + 2)M + 3Using L’hopital’s rule, we find that limM→∞ln(M+2)M+3= 0, so we have∞Xn=1ln(n + 1)n + 2−ln(n + 2)n + 3= limM→∞SM=ln 23.(9) Consider the following sequences. Circle the correct answers belowand show all supporting work.(I)n(−1)nn2− 12n2+ 1o∞n=1(II)nln(n)no∞n=1(III)nln1no∞n=1Sequence (I) (converges / diverges )Solution. Recall the following theorem: Suppose {an}∞n=1is a se-quence of nonnegative numbers. If {an}∞n=1converges to 0, then thesequence {(−1)nan}∞n=1also converges to 0. Otherwise, {(−1)nan}∞n=1diverges. We haven2− 12n2+ 1≥ 07for n ≥ 1. Sincelimn→∞n2− 12n2+ 1=12is nonzero, it follows from the theorem that the sequence diverges.Sequence (II) ( converges / diverges)Solution. The limit limx→∞ln(x)xis an indeterminate form of type∞∞. By L’hopital’s rule, we havelimx→∞ln(x)x= limx→∞1xx= limx→∞1x2= 0.Thereforelimn→∞ln(n)n= 0,so the sequence converges.Sequence (III) (converges / diverges )Solution. We have limn→∞1n= 0. Since limx→0+ln(x) = −∞, thisshows thatlimn→∞ln1n= −∞,so the sequence diverges.(10) Consider the following sequences. Circle the correct answers belowand show all supporting work.(I)n(−1)nn2− 12no∞n=1(II)n(−1)nn2− 12n2o∞n=1(III)n(−1)nn ln(n)o∞n=1Sequence (I) ( converges / diverges)Solution. We haven2− 12n≥ 08for n ≥ 1. The limit limx→∞x2−12xis indeterminate of type∞∞. ByL’hopital’s rule, we havelimx→∞x2− 12x= limx→∞2xln(2)2x.The right-hand limit is also indeterminate of the same type. Apply-ing L’hopital’s rule again, we getlimx→∞2xln(2)2x= limx→∞2ln(2)22x= 0.Thereforelimn→∞n2− 12n= 0,so the sequence converges by the theorem mentioned in the solutionto (9) part (I).Sequence (II) (converges / diverges )Solution. We haven2− 12n2≥ 0for n ≥ 1. Sincelimn→∞n2− 12n2=12is nonzero, it follows that the sequence diverges.Sequence (III) (converges / diverges )Solution. We haven ln(n) ≥ 0for n ≥ 1. Since {n ln(n)}∞n=1diverges to ∞, it does not converge to0, so the sequence of interest diverges.9(11) Consider the following series. Circle the correct answers below andshow all supporting work.(I)∞Xn=1n2+ 14n3− 1n(II)∞Xn=12n(n!)(n2)Series (I) ( converges / diverges)Solution. Each term of this series is an n-th powers whose basedepends on n. Therefore we should use the root test. We havelimn→∞n2+ 14n3− 1n1/n= limn→∞n2+ 14n3− 1= 0,so this series converges by the root test.Series (II) ( converges / diverges)Solution. The presence of the factorial is a good sign that we shoulduse the ratio test. We havelimn→∞2n+1((n+1)!)((n+1)2)2n(n!)(n2)= limn→∞2n+12n·1(n+1)!1n!·1(n+1)21n2= limn→∞2 ·n!(n + 1)!·n2(n + 1)2= limn→∞2 ·1n + 1nn + 12= 0since limn→∞1n+1= 0 and limn→∞nn+12= 1. Therefore the seriesconverges by the ratio test.(12) Consider the


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