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Tests for Convergence of Series 1) Use the Comparison Test to Confirm

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Tests for Convergence of Series 1) Use the Comparison Test to Confirm Tests for Convergence of Series 1) Use the comparison test to confirm the statements in the following exercises. P1 1 P1 1 1. n=4 n diverges, so n=4 n−3 diverges. Answer: Let an = 1=(n − 3), for n ≥ 4. Since n − 3 < n, we have 1=(n − 3) > 1=n, so 1 a > : n n P1 1 P1 1 The harmonic series n=4 n diverges, so the comparison test tells us that the series n=4 n−3 also diverges. P1 1 P1 1 2. n=1 n2 converges, so n=1 n2+2 converges. 2 2 2 2 2 Answer: Let an = 1=(n + 2). Since n + 2 > n , we have 1=(n + 2) < 1=n , so 1 0 < a < : n n2 P1 1 P1 1 The series n=1 n2 converges, so the comparison test tells us that the series n=1 n2+2 also converges. P1 1 P1 e−n 3. n=1 n2 converges, so n=1 n2 converges. −n 2 −n e−n 1 Answer: Let an = e =n . Since e < 1, for n ≥ 1,we have n2 < n2 , so 1 0 < a < : n n2 P1 1 P1 e−n The series n=1 n2 converges, so the comparison test tells us that the series n=1 n2 also converges. 2) Use the comparison test to determine whether the series in the following exercises converge. P1 1 1. n=1 3n+1 n n n n n 1 n Answer: Let an = 1=(3 + 1). Since 3 + 1 > 3 , we have 1=(3 + 1) < 1=3 = 3 , so 1n 0 < a < : n 3 P1 1 P1 1 n Thus we can compare the series n=1 3n+1 with the geometric series n=1 3 : This geometric series P1 1 converges since j1=3j < 1, so the comparison test tells us that n=1 3n+1 also converges. P1 1 2. n=1 n4+en 4 n 4 n 4 Answer: Let an = 1=(n + e ). Since n + e > n , we have 1 1 < ; n4 + en n4 so 1 0 < a < : n n4 P1 1 P1 1 Since the p-series n=1 n4 converges, the comparison test tells us that the series n=1 n4+en also converges. 1 X 1 3. ln n n=2 Answer: Since ln n ≤ n for n ≥ 2, we have 1= ln n ≥ 1=n, so the series diverges by comparison with the harmonic series, P 1=n. P1 n2 4. n=1 n4+1 2 4 4 4 1 1 Answer: Let an = n =(n + 1). Since n + 1 > n , we have n4+1 < n4 , so n2 n2 1 a = < = ; n n4 + 1 n4 n2 therefore 1 0 < a < : n n2 P1 1 P1 n2 Since the p-series n=1 n2 converges, the comparison test tells us that the series n=1 n4+1 converges also. P1 n sin2 n 5. n=1 n3+1 Answer: We know that j sin nj < 1, so n sin2 n n n 1 ≤ < = : n3 + 1 n3 + 1 n3 n2 P1 1 P1 n sin2 n Since the p-series n=1 n2 converges, comparison gives that n=1 n3+1 converges. P1 2n+1 6. n=1 n2n−1 n n n n n Answer: Let an = (2 + 1)=(n2 − 1). Since n2 − 1 < n2 + n = n(2 + 1), we have 2n + 1 2n + 1 1 > = : n2n − 1 n(2n + 1) n P1 2n+1 P1 1 Therefore, we can compare the series n=1 n2n−1 with the divergent harmonic series n=1 n : The comparison P1 2n+1 test tells us that n=1 n2n−1 also diverges. 3) Use the ratio test to decide if the series in the following exercises converge or diverge. P1 1 1. n=1 (2n)! Answer: Since an = 1=(2n)!, replacing n by n + 1 gives an+1 = 1=(2n + 2)!. Thus 1 jan+1j (2n+2)! (2n)! (2n)! 1 = 1 = = = ; janj (2n)! (2n + 2)! (2n + 2)(2n + 1)(2n)! (2n + 2)(2n + 1) so ja j 1 L = lim n+1 = lim = 0: n!1 janj n!1 (2n + 2)(2n + 1) P1 1 Since L = 0, the ratio test tells us that n=1 (2n)! converges. P1 (n!)2 2. n=1 (2n)! 2 2 Answer: Since an = (n!) =(2n)!, replacing n by n + 1 gives an+1 = ((n + 1)!) =(2n + 2)!. Thus, ((n+1)!)2 ja j ((n + 1)!)2 (2n)! n+1 = (2n+2)! = · : (n!)2 2 janj (2n + 2)! (n!) (2n)! However, since (n + 1)! = (n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have 2 2 2 jan+1j (n + 1) (n!) (2n)! (n + 1) n + 1 = 2 = = ; janj (2n + 2)(2n + 1)(2n)!(n!) (2n + 2)(2n + 1) 4n + 2 so ja j 1 L = lim n+1 = : n!1 janj 4 P1 (n!)2 Since L < 1, the ratio test tells us that n=1 (2n)! converges. P1 (2n)! 3. n=1 n!(n+1)! Answer: Since an = (2n)!=(n!(n + 1)!), replacing n by n + 1 gives an+1 = (2n + 2)!=((n + 1)!(n + 2)!). Thus, (2n+2)! ja j (2n + 2)! n!(n + 1)! n+1 = (n+1)!(n+2)! = · : (2n)! janj (n + 1)!(n + 2)! (2n)! n!(n+1)! However, since (n + 2)! = (n + 2)(n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have ja j (2n + 2)(2n + 1) 2(2n + 1) n+1 = = ; janj (n + 2)(n + 1) n + 2 so ja j L = lim n+1 = 4: n!1 janj P1 (2n)! Since L > 1, the ratio test tells us that n=1 n!(n+1)! diverges. P1 1 4. n=1 rnn! ; r > 0 n n+1 Answer: Since an = 1=(r n!), replacing n by n + 1 gives an+1 = 1=(r (n + 1)!). Thus 1 n jan+1j rn+1(n+1)! r n! 1 = 1 = n+1 = ; janj rnn! r (n + 1)! r(n + 1) so ja j 1 1 L = lim n+1 = lim = 0: n!1 janj r n!1 n + 1 P1 1 Since L = 0, the ratio test tells us that n=1 rnn! converges for all r > 0. P1 1 5. n=1 nen n n+1 Answer: Since an = 1=(ne ), replacing n by n + 1 gives an+1 = 1=(n + 1)e . Thus 1 n jan+1j (n+1)en+1 ne n 1 = 1 = n+1 = : janj nen (n + 1)e n + 1 e Therefore ja j 1 L = lim n+1 = < 1: n!1 janj e P1 1 Since L < 1, the ratio test tells us that n=1 nen converges. P1 2n 6. n=0 n3+1 n 3 n+1 3 Answer: Since an = 2 =(n + 1), replacing n by n + 1 gives an+1 = 2 =((n + 1) + 1). Thus n+1 2 n+1 3 3 jan+1j (n+1)3+1 2 n + 1 n + 1 = 2n = 3 · n = 2 3 ; janj n3+1 (n + 1) + 1 2 (n + 1) + 1 so ja j L = lim n+1 = 2: n!1 janj P1 2n Since L > 1 the ratio test tells us that the series n=0 n3+1 diverges. 4) Use the integral test to decide whether the following series converge or diverge. 1 X 1 1. n3 n=1 Answer: We use the integral test with f(x) = 1=x3 to determine whether this series converges or diverges. Z 1 1 We determine whether the corresponding improper integral 3 dx converges or diverges: 1 x Z 1 Z b b 1 1 −1 −1 1 1 3 dx = lim 3 dx = lim 2 = lim 2 + = : 1 x b!1 1 x b!1 2x 1 b!1 2b 2 2 1 Z 1 1 X 1 Since the integral dx converges, we conclude from the integral test that the series converges. x3 n3 1 n=1 1 X n 2. n2 + 1 n=1 Answer: We use the integral test with f(x) = x=(x2 +1) to determine whether this series converges or diverges. Z 1 x We determine whether the corresponding improper integral 2 dx converges or diverges: 1 x + 1 Z 1 Z b b x x 1 2 1 2 1 2 dx = lim 2 dx = lim ln(x + 1) = lim ln(b + 1) − ln 2 = 1: 1 x + 1 b!1 1 x + 1 b!1 2 1 b!1 2 2 1 Z 1 x X n Since the integral dx diverges, we conclude from the integral test that the series diverges. x2 + 1 n2 + 1 1 n=1 1 X 1 3. en n=1 Answer : We use the integral test with f(x) = 1=ex to determine whether this series converges or diverges. Z 1 1 To do so we determine whether the corresponding improper integral x dx converges or diverges: 1 e Z 1 Z b b 1 −x −x −b −1 −1 x dx = lim e dx = lim −e = lim −e + e = e : 1 e b!1 1 b!1 1 b!1 1 Z 1 1 X 1 Since the integral dx converges, we conclude from the integral test that the series converges. ex en 1 n=1 We can also observe that this is a geometric series with ratio x = 1=e < 1, and hence it converges. 1 X 1 4. n(ln n)2 n=2 Answer: We use the integral test with f(x) = 1=(x(ln x)2) to determine whether this series converges or Z 1 1 diverges.
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