This text explains the convergence and radius of convergence of different series using various tests such as comparison test, alternating series test, ratio test, and integral test. It also provides solved examples for each test. The text covers series with polynomial, exponential, and trigonometric functions.
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Order Id : 814869 Question 1: Test for convergence: i)∑ n=1 ∞1 n25nComparison test ∑ n=1 ∞1 n25n=∑ n=1 ∞1 n2∗∑ n=1 ∞1 5n Using the p-series ∑ n=1 ∞1 n2Converges since∑ n=1 ∞1 npand p>1 ∑ n=1 ∞1 5nConverges Therefore∑ n=1 ∞1 n25nconverges ii)∑ n=1 ∞n (n2+2)3comparison Test ∑ n=1 ∞n (n2+2)3Written in form of∑an∑bnand are two positive sequence: if∑bn converges so does∑an ∑ n=1 ∞n (n2+2)3<∑ n=1 ên n6And when factored, then∑ n=1 ê1 n5 Using the p-series∑ n=1 ê1 n5converges Thus,∑ n=1 ∞n (n2+2)3converges iii)∑ n=1 ∞n enIntegral Test
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f(x)=n enIf∫ n=1 ∞ f(x)dxconverges, so does∑ n=1 ∞n en ∫ n=1 ∞ f(x)dx=∫ n=1 ∞ n endn=2 e, so it converges at2 e Therefore,∑ n=1 ∞n enconverges iv)∑ n=1 ∞(−1)n+1 √nAlternating series test Suppose that foranthere exist an N so that for alln≥N anis positive and monotone decreasing andlim n→∞ an=0then the alternating series ∑(−1)n naand∑(−1)n−1anboth converges an=1 √nis positive and monotone decreasing lim n→∞(1 √n)=0then by Alternating series test criteria∑ n=1 ∞(−1)n+1 √nconverges v)∑ n=1 ∞ (−1)n+1n n+1Alternating series test Suppose that foranthere exist an N so that for alln≥N anis positive and monotone decreasing andlim n→∞ an=0then the alternating series ∑(−1)n naand∑(−1)n−1anboth converges an=n n+1is positive and not-monotone decreasing lim n→∞(n n+1)≠0Then by Alternating series test criteria∑ n=1 ∞ (−1)n+1n n+1diverges Question 2: Determining the radius and interval of convergence
i)∑ n=0 ∞xn 7n lim n→∞(|(xn 7n)1 n |):|x 7| The sum converges for L<1, solving for|x 7|<1 The radius of convergence isR=7 |x 7|<1Thus the interval is−7<x<7 The sum may converge for L=1, therefore the convergence for|x 7|=1 For x=7,∑ n=0 ∞7n 7n: diverges For x=-7,∑ n=0 ∞(−7)n 7n: diverges Therefore, the convergence interval∑ n=0 ∞xn 7nis−7<x<7 ii)∑ n=0 ∞(x+3)n 5n lim n→∞(|((x+3)n 5n)1 n |):|x+3 5| The sum converges for L<1, solving for|x+3 5|<1 The radius of convergence isR=5 |x+3 5|<1Thus the interval is−8<x<2
The sum may converge for L=1, therefore the convergence for|x+3 5|=1 For x=2,∑ n=0 ∞(2+3)n 5n: diverges For x=-8,∑ n=0 ∞((−8)+3)n 5n: diverges Therefore, the convergence interval(x+3)n 5nis−8<x<2 iii)∑ n=1 ∞(−1)n n(x−π)n Using the ratio test an+1 an = (−1)n+1 n+1(x−π)n+1 (−1)n n(x−π)n Computing the limit lim n→∞(|an+1 an|)=lim n→∞ (|(−1)n+1 n+1(x−π)n+1 (−1)n n(x−π)n|):|x−π| The sum converges for L<1, solving for|x−π|<1 The radius of convergence isR=1 |x−π|<1Thus the interval isπ−1<x<π+1 Forx=π+1:∑ n=1 ∞(−1)n n((π+1)−π)n converges (conditionally) Forx=π−1:∑ n=1 ∞(−1)n n((π−1)−π)ndiverges
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Therefore, the convergence interval(x+3)n 5nisπ−1<x≤π+1 iv)5 2+8(x−1)2 5 2+8(x−1)2Can be written as∑ n=0 ∞ 1 2(xn 5n) Now using the ratio test: an+1 an = 1 2(xn+1 5n+1) 1 2(xn 5n) Computing the limit lim n→∞(|an+1 an|)=lim n→∞ (|1 2(xn+1 5n+1) 1 2(xn 5n)|):|x 5| The sum converges for L<1, solving for|x 5|<1 The radius of convergence isR=5 |x 5|<1Thus the interval is−5<x<5 Forx=5:∑ n=0 ∞ ∑ n=0 ∞ 1 2(5n 5n)diverges Forx=−5:∑ n=0 ∞1 2((−5)n 5n)diverges Therefore, the convergence interval of5 2+8(x−1)2is−5<x<5
v)tan−1 (x 3) We know thattan−1(u)=∑ n=0 ∞(−1)n 2n+1u2n+1 Replace u withx 3gives∑ n=0 ∞(−1)n 2n+1(x 3) 2n+1 Now using the ratio test: an+1 an = (−1)n+1 2(n+1)+1(x 3) 2(n+1)+1 (−1)n 2n+1(x 3) 2n+1 Computing the limit lim n→∞(|an+1 an|)=lim n→∞ (|(−1)n+1 2(n+1)+1(x 3) 2(n+1)+1 (−1)n 2n+1(x 3) 2n+1 |):|x2 9| The sum converges for L<1, solving for|x2 9|<1 The radius of convergence isR=3 |32 9|<1Thus the interval is−3<x<3 Forx=3:∑ n=0 ∞(−1)n 2n+1(3 3) 2n+1 converges (conditionally) Forx=−3:∑ n=0 ∞(−1)n 2n+1(−3 3) 2n+1 converges (conditionally) Therefore, the convergence interval oftan−1 (x 3)is−3≤x≤3
vi)2x (1−x2)2 2x (1−x2)2Can be written as∑ n=0 ∞ (2n+2)x2n+1 Now using the ratio test: an+1 an =(2(n+1)+2)x2(n+1)+1 (2n+2)x2n+1 Computing the limit lim n→∞(|an+1 an|)=lim n→∞(|(2(n+1)+2)x2(n+1)+1 (2n+2)x2n+1|):|x|2 The sum converges for L<1, solving for|x|2<1 The radius of convergence isR=1 |x| 2<1Thus the interval is−1<x<1 Forx=1:∑ n=0 ∞ (2n+2)12n+1diverges Forx=−1:∑ n=0 ∞ (2n+2)(−1)2n+1diverges Therefore, the convergence interval of2x (1−x2)2is−1<x<1